Bai Tap Toan Roi Rac Truong Dh Su Pham Ky Thuat Hung Yen

TRNG I HC S PHM K THUT HNG YN

KHOA CNG NGH THNG TIN

BI TP

HC PHN TON RI RC 2

Trnh o to

H o to

:

:

i hc

Chnh quy/Lin thng

Bi tp TON RI RC 2 B mn Cng ngh phn mm - 2010

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LI NI U

C th ni ton hc ri rc l mn tin quyt v hiu qu nht ngi hc

nng cao t duy ton hc trong phn tch, thit k thut ton v rn luyn k nng

lp trnh vi nhng thut ton phc tp. Khng nhng th n cn l ca ng

ngi hc c th tip cn vi rt nhiu modul trong khoa hc my tnh (nh

Chng trnh dch, l thuyt tnh ton, Tr tu nhn to,...). Bi tp cng c

v nng cao kin thc trong mn hc ny

V ni dung, bm st vi chng trnh ca nh trng v h thng bi tp

cng c bin son theo cc chng l thuyt. Vi mi chng s c chia thnh

4 phn:

Phn A. Nhc li l thuyt: tm tt cc kin thc c bn, cc v d v cc

lu hu ch, cc kinh nghim trong khi lp trnh

Phn B. bi tp: a ra cc loi bi tp khc nhau, vi cc mc khc

nhau.

Phn C. Bi tp mu: Hng dn gii mt s bi tiu biu trong phn B, c

phn tch thut ton v ci t chng trnh.

Phn D. Bi tp t gii: Ngi hc thc hin vic gii cc bi tp ny

Mong rng ti liu ny p ng c phn no nhu cu ca hc sinh, sinh vin. y

l bn u tin chc chn cn rt nhiu sai st. Nhm tc gi mong nhn c s

ng gp ca cc thy c gio, cc bn sinh vin v ca tt c nhng ai quan tm ti

lnh vc ny.

Hng Yn, thng 7 nm 2010

B mn Cng ngh phn mm

Khoa Cng ngh thng tin

Trng i hc s phm k thut Hng Yn


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MC LC Bi 1: Cc khi nim c bn ca L thuyt th ............................................................5

Mc tiu ...................................................................................................................5 a. Nhc li l thuyt ..................................................................................................5 b. bi tp..............................................................................................................5 c. Hng dn gii......................................................................................................6 d. Bi tp t gii .......................................................................................................7

Bi 2: Biu din th trn my tnh..............................................................................10 Mc tiu .................................................................................................................10 a. Nhc li l thuyt ................................................................................................10 b. bi tp............................................................................................................10 c. Hng dn gii....................................................................................................10 d. Bi tp t gii .....................................................................................................14

Bi 3: th Euler .........................................................................................................15 Mc tiu .................................................................................................................15 a. Nhc li l thuyt ................................................................................................15 b. bi tp............................................................................................................16 c. Hng dn gii....................................................................................................16 d. Bi tp t gii .....................................................................................................19

Bi 4: th hamilton....................................................................................................20 Mc tiu .................................................................................................................20 a. Nhc li l thuyt ................................................................................................20 b. bi tp............................................................................................................20 c. Hng dn gii....................................................................................................20 d. Bi tp t gii .....................................................................................................22

Bi 5: Tho lun ci t th, cc thut ton lit k chu trnh Euler v Hamilton. Tho lun v bi tp ln.................................................................................................23

Mc tiu .................................................................................................................23 a. Nhc li l thuyt ................................................................................................23 b. bi tp............................................................................................................23 c. Hng dn gii....................................................................................................23 d. Bi tp t gii .....................................................................................................31

Bi 6 Thut ton tm kim trn th v ng dng.........................................................34 Mc tiu .................................................................................................................34 a. Nhc li l thuyt ................................................................................................34 b. bi tp............................................................................................................34 c. Hng dn gii....................................................................................................34 d. Bi tp t gii .....................................................................................................51

Bi 7: Cy v cy khung ................................................................................................52 Mc tiu .................................................................................................................52 a. Nhc li l thuyt ................................................................................................52 b. bi tp............................................................................................................53 c. Hng dn gii....................................................................................................54 d. Bi tp t gii .....................................................................................................55

Bi 8: Tho lun v ci t thut ton tm cy khung nh nht trn th ......................58 Mc tiu .................................................................................................................58


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a. Nhc li l thuyt ................................................................................................58 b. bi tp............................................................................................................58 c. Hng dn gii....................................................................................................58 d. Bi tp t gii .....................................................................................................70

Bi 9, 10: Bi ton tm ng i ngn nht ....................................................................71 Mc tiu .................................................................................................................71 a. Nhc li l thuyt ................................................................................................71 b. bi tp............................................................................................................71 c. Hng dn gii....................................................................................................73 d. Bi tp t gii .....................................................................................................92

Bi 12: Bi ton lung cc i trong mng.....................................................................97 Mc tiu .................................................................................................................97 a. Nhc li l thuyt ................................................................................................97 b. bi tp............................................................................................................98 c. Hng dn gii....................................................................................................99 d. Bi tp t gii ...................................................................................................101


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Bi 1: Cc khi nim c bn ca L thuyt th Mc tiu

- Lu tr c th trn my tnh theo nhng phng php khc nhau.

- Ci t c chng trnh chuyn i gia cc phng php.

- Sinh vin c kh nng t hc.

a. Nhc li l thuyt

- Hai nh x, y c gi l cp nh lin thng , nu hoc gia x v y c t nht mt

xch ni vi nhau, hoc tn ti t nht mt ng i t y sang x.

- th v hng G(V,E) c gi l th lin thng, nu mi cp nh ca n

u lin thng.

- th c hng G(V,E) c gi l th lin thng mch, nu mi cp nh ca

n u lin thng.

- Biu din dng hnh hc: Gi s c th G(V,E).

Biu din nh: ly cc im trn mt phng hay trn khng gian tng ng

vi cc phn t ca tp V v dng ngay k hiu cc phn t ny ghi trn cc

im tng ng.

Biu din cnh: Nu cnh a vi hai nh u l x,y th n c biu din

bng on thng hay mt on cong ni gia hai im x, y v khng i qua cc

im tng ng trong khng gian.

Biu din cung: nu cung a c nh u l x, nh cui l y, th n c biu

din bng mt on thng hoc on cong c nh hng i t x sang y v khng

qua cc im tng ng trung gian khc.

Hnh nhn c gi l dng biu din hnh hc ca th G(V, E). i khi

ngi ta cng gi dng biu din hnh hc l mt th.

b. bi tp

Bi 1 Cho G th gm 4 phn G1, G2, G3 v G4 nh sau:

a. Ch ra tp nh, cnh(v hng,c hng, khuyn,..) ca mi th cho? Ch

loi th ?

b. th G, G1, G2, G3, G4 v G5 c lin thng ko? Nu th ko lin thng hy

ch ra cc thnh phn lin thng?


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c. th G, G1, G2, G3, G4 v G5 c chu trnh ko? Ch ra cc chu trnh ca th

(nu c)?

c. Hng dn gii

Bi 1

a.

Tn th Tp nh V Tp cnh E Loi th

G1 1,2,3,4 (1,2);(1,4);(2,3);(2,4);(3,4) V hng

G2 5,6,7 (5,6);(5,7);(6,7) C hng

G3 8,9 (8,9) V hng

G4 0

G 1,2,3,4,5,6,7,8,9,0 (1,2);(1,4);(2,3);(2,4);(3,4);

(8,9)

(5,6);(5,7);(6,7)

Hn hp

b.

Tn th Tnh lin thng Tn thnh phn lin thng

G1 C G1

G2 C G2

G1

1

4 3

2 5

7 6

8

9

00

G2 G3

G4


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G3 C G3

G4 C G4

G Khng G1,G2,G3,G4

c.

Tn th C chu trnh? Tn chu trnh

G1 C (1,2,4,1);(1,2,3,4,1);(2,3,4,2)

G2 Khng

G3 Khng

G4 Khng

G C (1,2,4,1);(1,2,3,4,1);(2,3,4,2)

d. Bi tp t gii

Bi 1. Mt qun o c n( n ) hn o v hai hn o bt k thuc qun o u

c s u mi ng ngm ti mt trong nhng hn o ny u nh hn n. Chng

minh rng t mt hn o ty thuc qun o ta c th i n mt hn o bt k

khc ca qun o bng ng ngm.

Bi 2 Khi v ngh h mi bn hc sinh ca lp 11A trng L Hng Phong u

trao i a ch vi t nht mt na s bn trong lp. Chng minh rng trong thi

gian ngh h mi bn ca lp 11A u c th bo tin trc tip hay gin tip cho cc

bn trong lp.

Bi 3 Trong mt cuc hp c ng hai i biu khng que nhau v mi i biu ny

c mt s l ngi que n d. Chng minh rng lun lun c th xp mt s i

bieetr ngi chen gia hai i bi ni trn , ngi ngi gia hai ngi m anh(

ch) ta quen.

Hng n:


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gii c bi ton trn trc ht ta xy dng cc th tng ng, sau vn

dng kt qu ca nh l 4.1, h qu 4.1 v nh l 4.2 m suy ra kt lun.

Xuy dng th

nh: Ly cc im trong mt phng hay trong khng gian tng ng vi cc

hn o thuc qun o ( cc bn hc sinh trong lp 11A, cc i biu n

hp).

Cnh: Hai im x, y c ni bng mt cnh khi v ch khi hai hn o x, y

c ng ngm trc tip vi nhau( cc bn x, y trao i a ch cho nhau, cc

i biu x, y quen nhau)

- th nhn c k hiu bng G1 , (G2 , G3)

- th G1 m t ton b li ng ngm trong qun o

- th G2 m t ton b quan h trao i a ch trong lp 11A

- th G3 m t ton b quen bit trong cc i biu trong cc i biu n

d hp.

Vn dng kt qu cc nh l suy ra kt lun

- Do hai hn o bt k u c tng s u mi ng ngm khng nh hn n, nn

hai nh bt k ca th G1 u c tng bc khng nh hn n. Bi vy theo nh l

4.1. th G1 lin thng, nn hai hn o bt k c ng hm ni vi nhau.

- V mi bn hc sinh trong lp 11A trao i a ch vi t nht mt na s bn tron

lp, nn bc ca mi nh ca G2 khng nh hn mt na s nh ca th. Khi

, theo h qu 4.1. th G2 lin thng. Bi vy hai nh x, y u c xch ni vi

nhau. Khi thng qua cc bn tng ng vi cc nh thuc xch , m bn tng

ng vi nh x bo tin c cho tng ng vi nh y v ngc li.

- Hai i biu khng quen nhau, th hai nh tng ng khng k nhau. Mi i biu

ny li c mt s l ngi quen n hp, nn trong th lin thng G3 c ng hai

nh bc l v hai nh ny li khng k nhau. Khi d, theo nh l 4.2, hai nh ny

lin thng nn c t nht mt xich ni gia hai nh ny. Gi s l mt trong

nhng mi xch ni gia hai bc l ny. Da vo ta sp xp cc i biu tng

ng ngi gia hai ngi m anh ch quen.

Bi 4 Cho G th nh sau:

Ch ra tp nh, cnh(v hng,c hng, khuyn,..) ca mi th cho? Ch

loi th ? th c lin thng ko? Nu th ko lin thng hy ch ra cc


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thnh phn lin thng? th c chu trnh ko? Ch ra cc chu trnh ca th (nu

c)?


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Bi 2: Biu din th trn my tnh Mc tiu

- Nu c cc cch biu din th v biu din th trn my tnh trn my tnh.

- a ra c ma trn k, danh sch cc cnh, cung tng ng vi 1 th cho

trc.


- - Phn tch c bi ton thc t tng ng phn l thuyt hc.


a. Nhc li l thuyt

b. bi tp

Bi 1 Cho G th gm 4 phn G1, G2, G3 v G4 nh sau:

a. Biu din cc th G,G1,G2,G3,G4 di dng ma trn k

b. Biu din cc th G,G1,G2,G3,G4 di dng danh sch cnh(cung)

c. Biu din cc th G,G1,G2,G3,G4 di dng danh sch k

Bi 2 Ci t chng trnh nhp danh sch k ca th t bn phm v a danh

sch ra mn hnh.

c. Hng dn gii

Bi 1

G1

1

4 3

2 5

7 6

8

9

00

G2 G3

G4


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Tn th a. ma trn k b. danh sch

cnh(cung)

c.danh sch k

G1 1 2 3 4

1 0 1 0 1

2 1 0 0 1

3 0 0 0 1

4 1 1 1 0

Danh sch cnh

1 2

1 4

2 3

2 4

3 4

1 2 4

2 1 4

3 4

4 1 2 3

G2 5 6 7

5 0 0 1

6 1 0 1

7 0 0 0

Danh sch cung

5 6

5 7

6 7

5 6 7

6 7

G3 8 9

8 0 1

9 1 0

Danh sch cnh

8 9

8 9

9 8

G4 0

0 0

G 1 2 3 4 5 6 7 8 9 0

1 0 1 0 1 0 0 0 0 0 0

2 1 0 0 1 0 0 0 0 0 0

3 0 0 0 1 0 0 0 0 0 0

4 1 1 1 0 0 0 0 0 0 0

5 0 0 0 0 0 0 1 0 0 0

Danh sch cung

1 2

1 4

2 1

2 3

2 4

1 2 4

2 1 4

3 4

4 1 2 3

5 6 7

6 7


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6 0 0 0 0 1 0 1 0 0 0

7 0 0 0 0 0 0 0 0 0 0

8 0 0 0 0 0 0 0 0 1 0

9 0 0 0 0 0 0 0 1 0 0

0 0 0 0 0 0 0 0 0 0 0

3 2

3 4

4 1

4 2

4 3

5 6

5 7

6 7

8 9

9 8

0

Bi 2 Chng trnh nhp danh sch k ca th t bn phm v a danh sch

ra mn hnh

Phn tch bi ton :

Trong rt nhiu thut ton lm vic vi th chng ta thng xuyn phi thc hin

cc thao tc: Thm hoc bt mt s cnh. Trong trng hp ny Cu trc d liu

dng trn l khng thun tin. Khi nn chuyn sang s dng danh sch k lin

kt (Linked Adjancency List) nh m t trong chng trnh nhp danh sch k ca

th t bn phm v a danh sch ra mn hnh

Chng trnh minh ha :

Program AdjList;

Const

maxV=100;

Type

link=^node;

node=record


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v:integer;

next:link;

End;

Var

j,x,y,m,n,u,v:integer;

t:link;

Ke:array[1. .Vmax] of link;

Begin

Write(Cho so canh va dinh cua do thi:); readln(m,n);

(*Khoi tao*)

for j:=1 to n do Ke[j]:=nil;

for j:=1 to m do

begin

write(Cho dinh dau va cuoi cua canh ,j,:);

readln(x,y);

new(t); t^.v:=x, t^.next:=Ke[y]; Ke[y]:=t;

new(t); t^.v:=y, t^.next:=Ke[x]; Ke[x]:=t;

end;

writeln(Danh sach ke cua cac dinh cua do thi:);

for J:=1 to m do

begin

writeln(Danh sachcac dinh ke cua dinh ,j,:);

t:=Ke[j];


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while t^.nextnil do

begin

write(t^.v:4);

t:=t^.next;

end;

end;

readln;

End.

d. Bi tp t gii


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Bi 3: th Euler Mc tiu

- Kim tra c mt th bt k c l th euler hay khng.

- p dng c thut ton tm chu trnh Euler, ng Euler, vi 1 th cho trc.

- Lu tr c th trn my tnh theo nhng phng php khc nhau. Ci t

c chng trnh chuyn i gia cc phng php.

- Ci t c thut ton Tm chu trnh Euler.

- Ci t c thut ton duyt th duyt theo chiu su hoc duyt theo chiu

rng.


a. Nhc li l thuyt

- Chu trnh (t.. ng i) n cha tt c cc cnh (hoc cung) ca th (v

hng hoc c hng) G c gi l chu trnh (t.. ng i) Euler. Mt th

lin thng (lin thng yu i vi th c hng) c cha mt chu trnh (t..

ng i) Euler c gi l th Euler (t.. na Euler).


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nh l

th (v hng) lin thng G l th Euler khi v ch khi mi nh ca G u c

bc chn.

H qu

th lin thng G l na Euler (m khng l Euler) khi v ch khi c ng hai

nh bc l trong G.

Thut ton vch c mt chu trnh Euler trong th lin thng G c bc ca mi

nh l chn theo thut ton Fleury sau y.

Xut pht t mt nh bt k ca G v tun theo hai quy tc sau:

1. Mi khi i qua mt cnh no th xo n i; sau xo nh c lp (nu c);

2. Khng bao gi i qua mt cu, tr phi khng cn cch i no khc.

b. bi tp

Bi 1: th sau c l th na Euler hay th Euler ko? Gii thch?

c. Hng dn gii

Bi 1: th sau c l th na Euler hay th Euler ko? Gii thch?


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th G1 l th na Euler

Tht vy, cc nh a,b,e c bc l 2,4,4 l bc chn, cc nh c v d c bc l 3 l

bc l

th G2 l th na Euler

Tht vy, cc nh c v d c bc l 2 l bc chn, cc nh a v c c bc l 1 l bc

l

a

b

e

c

d

G1

a

d

b

c

G2


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th G3 khng l th Euler

Tht vy G3 c 3 nh a,d v g c bc l 1 l bc l

G4

th G4 l th na Euler v theo h qu..

Tht vy, cc nh 3,4,5 c bc l 2 l bc chn, cc nh 1 v 2 c bc l 3 l bc l

Bi 2

g

a

d

b

c

e

f

G3

1

2

3 4

5


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Xut pht t u, ta c th i theo cnh (u,v) hoc (u,x), gi s l (u,v) (xo

(u,v)). T v c th i qua mt trong cc cnh (v,w), (v,x), (v,t), gi s (v,w) (xo

(v,w)). Tip tc, c th i theo mt trong cc cnh (w,s), (w,y), (w,z), gi s (w,s)

(xo (w,s)). i theo cnh (s,y) (xo (s,y) v s). V (y,x) l cu nn c th i theo mt

trong hai cnh (y,w), (y,z), gi s (y,w) (xo (y,w)). i theo (w,z) (xo (w,z) v w)

v theo (z,y) (xo (z,y) v z). Tip tc i theo cnh (y,x) (xo (y,x) v y). V (x,u) l

cu nn i theo cnh (x,v) hoc (x,t), gi s (x,v) (xo (x,v)). Tip tc i theo cnh

(v,t) (xo (v,t) v v), theo cnh (t,x) (xo cnh (t,x) v t), cui cung i theo cnh

(x,u) (xo (x,u), x v u).

d. Bi tp t gii


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Bi 4: th hamilton Mc tiu

- Kim tra c mt th bt k c l th Hamilton hay khng.

- p dng c thut ton tm chu trnh Hamilton, ng Hamilton, vi 1 th

cho trc.

- Lu tr c th trn my tnh theo nhng phng php khc nhau. Ci t

c chng trnh chuyn i gia cc phng php.

- Ci t c thut ton Tm chu trnh Halmiton.

- Ci t c thut ton duyt th duyt theo chiu su hoc duyt theo chiu

rng.


a. Nhc li l thuyt

- th Hamilton(na Hamilton) l th c cha mt chu trnh(ng i)

Hamilton

Hay

ng i (x[1],x[2],,x[n]) c gi l ng i Hamilton nu x[i]x[j]

(1i


Trang 21

th G1 l th Hamilton

th G2 l th na Hamilton

th G3 khng c chu trnh hay ng i Hamilton

a

b

e

c

d

G1

a

d

b

c

G2

g

a

d

b

c

e

f

G3


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G4

th G4 l th Hamilton, v c chu trnh Hamilton (1,3,5,2,4,1)

d. Bi tp t gii

1

2

3 4

5


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Bi 5: Tho lun ci t th, cc thut ton lit k chu trnh Euler v Hamilton. Tho lun v bi tp ln Mc tiu


- Chuyn i gia cc kiu biu din th trn my tnh.

- Nng cao kh nng lm vic nhm.

- Rn luyn t duy sng to.

- Phn tch c bi ton thc t tng ng phn l thuyt hc.


a. Nhc li l thuyt

Xem li trong bi 3 v bi 4

b. bi tp

Bi 1 Tm chu trnh Euler trong th G.

Bi 2 Tm chu trnh Halmiton trong th G.

c. Hng dn gii

Bi 1 bi : Ci t chng trnh tm chu trnh Euler trong th G.

Cho th G=(X,E) tn ti chu trnh Euler. Hy tm chu trnh (chi trnh Euler l

chu trnh i qua tt c cc cnh ca th, mi cnh i qua ng mt ln).

Phn tch bi ton :

u vo: th G

u ra: Chu trnh Euler (nu c)

2. Thut ton:

Xut pht t mt nh u bt k, khi i qua cnh no th xo cnh khi th v

ghi li t tri sang phi. Khi thc hin thut ton cn lu nu gp cnh bc cu

gia 2 thnh phn lin thng th ta phi xo ht thnh phn lin thng ri mi xo

n cnh bc cu.


Trang 24

Khi xo cnh bc cu th phi loi ht cc nh tr tri (ngha l khng k vi bt c

nh no thuc th).

Cu trc d liu:

Biu din th bng ma trn k a(i,j), do lu khi xo i mt cnh ta ch vic

gn a(i,j)=a(j,i)=0, ng thi phi lu cnh va xo vo mt mng khc: Mng Ctr.

Mng lt: array [1Nmax] of integer dng trong th tc tm thnh phn lin thng

(ging nh mt thut ton tm thnh phn lin thng trnh by trn).

Mng dd: array [1Nmax] of Boolean, gi tr dd[i] cho bit nh i b loi khi

th hay cha. Nu b li th dd[i]=True; ngc li dd[i]=False;

Th tc

Procedure Euler_Cycle;

Begin

STACK:= ; CE:= ;

Chon u la mot dinh nao do cua do thi;

STACK u;

While STACK do

Begin

X:=top(STACK); (* x la phan tu dau STACK)

If Ke(x) then

Begin

Y:=dinh dau tien trong danh sach Ke(x);

STACK y;

(* loai bo canh (x,y) khoi do thi *)

Ke(x):=Ke(x)\ { y} ;

Ke(y):=Ke(y)\{ x} ;


Trang 25

End

Else

Begin

x STACK; CE x;

End;

End;

End;

Chng trnh minh ha :

Program Chu_trinh_Euler;

Const Nmax=100;

Emax=100;

Var a:array[1Nmax,1..Nmax]of integer;

Ctr: array[1Emax,1,2]of integer;

dd: array[1Nmax]of Boolean;

lt: array[1Nmax]of integer;

SolC,N:integer;

(*Bin Solc l s cnh ca chu trnh, s tng ln mt n v mi khi ta i qua

mt cnh*).

Program Timtplt(k:integer);

(* Th tc ny tm thnh phn lin thng ging nh thut ton tnh thnh phn

lin thng trnh by trn ch khc mt im l ta ch tm tplt i vi cc nh cha

b loi khi th *).

Var i:integer;

Begin lt[k]:=sotp;


Trang 26

For i=1 to N do

If dd[i] and (lt[i]=0) and(a[i,k]=1)then

Begin

lt[i]:=sotp;

Timtplt(i);

End;

End;

Function KTLT (u,v:integer):Boolean;

(* Hm ny kim tra xem khi ta xo cnh (u,v) i th th cn lin thng na hay

khng? Nu cn th s tplt lun bng 1. y coi nh kim tra cnh (u,v) c phi l

cnh bc cu hay khng*).

Var i:integer;

Begin(* u tin ta phi xo cnh (u,v) khi th ri mi tin hnh kim

tra *).

a[u,v]:=0;a[u,v]:=0;

Fillchar(lt,sizeof(lt),0);

(* on m sau tm s tplt ca th im ang xt *).

Sotp:=0;

For i=1 to N do

If dd[i] and (lt[i]=0)then

Begin

inc(sotp);

Timtplt(i);

End;


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If sotp=1 then KTLT=True

Else

Bengin (*nu khng c cnh bc cu th khi khi phc

li cnh (u,v)*).

a[u,v]:=1

a[v,u]:=1

KTLT:=False

End;

End;

Procedure Timchutrinh;

Var i,u,dem:integer;

Begin solc:=0;

u:=1;

(* Chu trnh ca ta xut pht t nh 1, tuy nhin iu ny khng bt buc *)

For i=1 to N do

dd[i] true;

(* Khi to tt c cc gi tr dd[i]=True ngha l cha c nh no b

loi *)

REPEAT

Dem:=0;

(* Bin m l s cnh k vi nh u *)

Begin

For i=1 to N do

If a[u,i]=1 then


Trang 28

Begin (*xo cnh (u,i) khi th *)

a[u,i]:=0

a[i,u]:=0

(*Loi nh u khi th*)

dd[u]:=False;

(*ua cnh (u,i) vo chu trnh*).

inc(solC);

Ctr[solC,1]:=u;

Ctr[solC,2]:=i;

Break;

End;

End

Else

(*Ngc li nu cn nhiu cnh ni vi u ta tm mt cnh khng phi l cnh

bc cu v xo cnh ng thi a n vo chu trnh*)

If dem>1 then

Begin

For i=1 to N do

If (a[u,i]=1 and dd[i] and KTLT(u,i)then

(*Nu cnh (u,i) tho mn khng phi l cnh bc cu th ta chn:

Ngha l n phi k vi u, cha b loi khi th v sau khi xo cnh k

th th vn lin thng. Ring iu kin th 3 ta dng hm KTLT(u,i)

kim tra xem khi xo cnh (u,i) i th th cn lin thng na hay khng*).

Begin

(*on m sau xo cnh (u,i) v a cnh ny


Trang 29

vo chu trnh*).

a[u,i]:=0

a[i,u]:=0

inc(solC);

Ctr[solC,1]:=u;

Ctr[solC,2]:=i;

Break;

(*Sau khi tm c nh i th ta phi thot khi vng lp For nu khng c

lnh Break th chng trnh s chy sai*).

End;

End;

(*Chun b cho php tip chu trnh ca chng ta li xut pht t nh*)

u:=1;

UNTIL dem=0;

End;

Procedure Ghifile;

Var f:text;

I:integer;

Begin

Assign(f,kq.out);Rewrite(f);

For i:=1 to solC do

writeln(f,Ctr[i,1], ,Ctr[i,2]);

Close(f);

BEGIN


Trang 30

Docfile;

Timchutrinh;

Ghifile;

END.

Bi 2 bi : Lit k tt c cc chu trnh Hamilton ca th

Phn tch bi ton :

Thut ton sau y c xy dng da trn c s thut ton quay lui cho php lit

k tt c cc chu trnh Hamilton ca th.

Hnh di y m t cy tm kim theo thut ton va m t.

th v cy lit k chu trnh Hamilton ca n theo thut ton quay lui.

Trong trng hp th c khng qu nhiu cnh thut ton trn c th s dng

kim tra th c phi l Hamilton hay khng.

Chng trnh minh ha :

Procedure Hamilton(k);

(* liet ke cac chu trinh Hamilton thu duoc bang viec phat trien day dinh (X[1],. . .

, X[k-1]) cua do thi G=(V,E) cho boi danh sach ke: Ke(v), v V *)

begin


Trang 31

for y Ke(X[k-1]) do

if (k =N+1) and (y=v0) then Ghinhan(X[1],. . . , X[n], v0)

else

if Chuaxet[y] then

begin

X[k]:=y;

Chuaxet[y]:=false;

Hamilton(k+1);

Chuaxet[y]:=true;

end;

end;

(* Main program*)

begin

for v V do Chuaxet[v]:=true;

X[1]:=0; (* v0 la mot dinh nao do cua do thi *)

Chuaxet[v0]:=false;

Hamilton(2);

end.

d. Bi tp t gii

Bi tp 1: Hi ngh bn trn

Tng th k i hi ng Lin hp quc triu tp mt cuc hp c N nh

ngoi giao ca N t chc tham gia. Cc i din ngoi giao c b tr ngi quanh

mt bn trn. Gia mt s t chc c quan h cng thng, v vy khng th xp h


Trang 32

ngi cnh nhau c. Thng tin v quan h gia cc t chc c cho di dng

cp s nguyn i, j nu gia 2 t chc ny c quan h cng thng.

Hy lp trnh gip Tng th k Lin hp quc b tr ch ngi quanh bn hp. Cc

t chc c nh s t 1 ti N, 0 < N


Trang 33

qut, hy sp N ngi ngi chung quanh bn trn sao cho mi ngi u ngi cnh

ngi mnh quen. Bit mi ngi c t nht (N + 1)/2 ngi quen.


Trang 34

Bi 6 Thut ton tm kim trn th v ng dng Mc tiu

- Trnh by c tng, cch ci t v ci t c thut ton BFS, DFS.

- Nu u nhc ca tng thut ton i vi tng loi th.




a. Nhc li l thuyt

b. bi tp

Bi 1 Dng cy m t kt qu duyt chiu su v duyt theo chiu rng trn

th sau:

Bi 2 Cho th G=(X,E) vi tp cc nh X v tp cc cung E. Xt xem th c

bao nhiu thnh phn lin thng, mi thnh phn lin thng bao gm nhng nh

no?

Bi 3 Thut ton tm ng i theo chiu su (thut ton duyt theo chiu su)

Bi 4 Thut ton tm ng i theo chiu rng (thut ton duyt theo chiu rng)

c. Hng dn gii

Bi 1 Dng cy m t kt qu duyt chiu su v duyt theo chiu rng trn

th sau:


Trang 35

Duyt rng

Duyt su:

Bi 2 Thut ton tm s thnh phn lin thng

Cho th G=(X,E) vi tp cc nh X v tp cc cung E. Xt xem th c

bao nhiu thnh phn lin thng, mi thnh phn lin thng bao gm nhng nh

no?

Phn tch bi ton :

3 5


Trang 36

8 2 6 3

1

2

2 7 9 2

1 1 3 1

4 1

th vi cc thnh phn LT

u vo:

a. Biu din th bng ma trn k

Dothi.txt Kq.txt

9

0 1 1 1 0 0 0 0 0

1 0 1 0 0 0 0 0 0

1 1 0 1 0 0 0 0 0

1 0 1 0 0 0 0 0 0

0 0 0 0 0 0 0 1 0

0 0 0 0 0 0 1 0 1

So tph cua do thi la:3

TPLT 1 bao gom cac dinh: 1 2 3 4

TPLT 2 bao gom cac dinh: 5 8

TPLT 3 bao gom cac dinh: 6 7 9


Trang 37

0 0 0 0 0 1 0 0 1

0 0 0 0 1 0 0 0 0

0 0 0 0 0 1 1 0 0

Cu trc d liu

u tin nh s cc nh l 0

(Dng mt mng biu din cc thnh phn l mng chng ta cn tm, ban u

ta gn tt c cc gi tr ca mng bng 0)

Ly ra nh L gn lt[i]:=1

+ Duyt cc nh k ca 1 gp nh tip theo l nh 2 cha c nh s ta

nh s 1 gn lt[2]:=1 sang bc tip theo.

+ Duyt cc nh k ca 2, gp nh 1 c nh s ri nn b qua, gp nh

cha nh s nn nh s 1, gn lt[4]:=1 sang bc tip theo.

+ Duyt cc nh k ca 4, gp nh 1 nh s ri nn b qua. Gp nh 2 nh

s ri nn b qua. Gp nh 3 cha nh s nn nh s 1, lt[3]:=1 sang bc tip

theo.

+ Duyt cc nh k ca 3 gp ton cc nh nh s ri nn khng nh s

na dng vic nh s.

Nh vy ta tm c 1 thnh phn lin thng.

Ly ra mt nh cha nh s v d ta ly nh 6

Ta li lm nh trn ta li tm c thnh phn lin thng khc.

Nh vy ta rt ra c Cu trc d liu ca bi ton ny.

i vi th v cc cnh th ta c th dng ma trn k hoc danh sch k, v

nn c t file n gin ho.

Dng mt mng lt[1n] biu din TPLT lt[i]:=K nu nh i thuc thnh phn

lin thng.

Chng trnh minh ha :


Trang 38

a. Biu din th bng ma trn k.

a(i,j)=1 nu 2 nh i v j k nhau v =0 nu 2 nh i v j khng k nhau.

Program tim_thanh_phan_lien_thong;

Const Nmax=100;

Type mang=array[1Nmax] of integer;

Var lt:mang;

a: array[1Nmax, 1Nmax] of byte;

N:integer;

Sotp: integer;

(*Th tc ny c t file dothi.txt s N l s nh ca th ma trn k biu din

th *)

Procedure Docfile;

Var i,j: integer;

f: text;

Begin

Assign(f,dothi.txt);reset(f);

Readln(f,N);

For i:=1 to N do

Read(f,a[i,j]);

Close(f);

End;

(*Th tc quy ny dng tm cc nh thuc mt thnh phn lin thng *)

Procedure Docfile;

Var i,j: integer;


Trang 39

Begin lt[k]:=sotp;

(*Duyt qua cc nh ca th *)

For i:=1 to N do

(*Nu nh i k vi k v cha nh du th ta gi th tc quy. Tm

LT ngha l li nh du v duyt cc nh k ca i *)

lf (lt[i]=0)and(a[i,k]=1)then

TimLT(i);

End;

(*Th tc ny c coi nh bc 2 trong v d *)

Procedure TimTPLT;

Var i: integer;

Begin sotp:=0;

(*Duyt cc nh cha c nh du *)

For i:=1 to N do

(*Nu c mt nh cha thuc thnh phn lin thng no lt[i]=0 ngha

l tm ra mt thnh phn mi nn ta tng s lng thnh phn v bt u ta i

tm thnh phn lin thng mi bng th tc quy TimLT(i)*)

lf lt[i]=0 then

Begin

Inc(sotp);

TimLT(i);

End;

End;

(*Th tc in ra cc thnh phn lin thng ca th *)


Trang 40

Procedure Ghifile;

Var i,j: integer;

Begin

Writeln(so thanh phan lien thong cua do thi l: ;sotp);

For i:=1 to sotp do

Begin

Writeln(TPLT, i ,bao gom cac dinh:);

For j:=1 to N do

(*nu j thuc thnh phn lin thng th i th in ra *)

lf lt[j]=i then

Write(J, );

Writeln;

End;

Readln;

End;

BEGIN

Docfile;

Tim TPLT;

Ghifile;

END.

b. Biu din th bng danh sch cnh

Ma trn ca chng ta ( v d trn) c vit di dng:

2 3 4 9


Trang 41

1 3 0 1 2 3 4

2 4 0 2 1 3

1 3 0 3 2 4

8 0 0 4 1 3

7 9 0 5 8

6 9 0 6 7 9

5 0 0 7 6 9

6 7 0 85

9 6 7

a[i,9] l s lng cc nh k vi nh i;

a[i,j] chnh l cc nh k vi cc nh i (j=1,2,3,)

Khi th tc Docfile c vit li nh sau:

Procedure Docfile;

Var i,j,t: integer;

f: text;

Begin Assign(f,dothi.txt);reset(f);

Readln(f,N);

For i:=1 to sotp do

Bengin

Read(f,t);

While not eoln(f) do

Begin

Read(f,t);


Trang 42

Inc(j);

a[i,j]:=t;

End;

a[i,0]:=j;

End;

Close(f);

End;

Th tc Ghifile v thuc tc TimTPLT khng c g thay i

Th tc TimTPLT thay i nh sau:

Procedure TimLT (k:integer);

Var i: integer;

Begin

Lt[k]:=sotp;

For i:=1 to a[k,0] do

If lt[a[k,i]]=0 then

TimLT(a[k,i]);

End;

Bi 3 bi : Thut ton tm ng i theo chiu su (thut ton duyt theo chiu

su).Tm ng i gia hai nh xp v kt trn th G v hng v khng c trng

s.

Phn tch bi ton :


Trang 43

Thut ton ca bi ton ny cng ging nh thut ton tm thnh phn lin

thng ca th. Trong phn Cu trc d liu c dng mng trc lu ng i

t nh xp n nh kt.

Chng trnh minh ha :

Program Duyetchieusau;

Const Nmax=100;

Type Mang=array[1Nmax] of integer;

Matran= array[1Nmax,1Nmax] of integer;

Var a:Matran;

so,duong:mang;

N:integer;

(*th tc c tp v ghi tp ging nh bi trc ch c thm hai nh xp v kt

*)

Procedure Duyet (k:integer);

Var i:integer;

Begin

For i:=1 to N do

(*Duyt cc nh k vi k *)

lf (a[i,k]=1)and(truoc[i]=0)then

Begin (*ta gn nh i trc nh i l nh k *)

Truoc[i]:=k;

Duyet(i);

End;

End;


Trang 44

Procedure Duyetsau;

Var i:integer;

Begin

For i:=1 to N do truoc[i]:=0

Truoc[xp]:=-1;

Duyet(xp);

End;

Procedure Timduong;

(*th tc ny ging th tc tm ngc trong duyt chiu rng, nhng ch tm

mt ng i ngn nht trong rt nhiu ng i ngn nht t xp n kt *).

Var i,u: integer;

Begin

sold:=0;

u:=kt;

REPEAT

Inc(sold);

Duong[sold]:=u;

U:=truoc[u];

UNTIL u=-1;

End;

BEGIN

Doctiep;

Duyetsau;

Timduong;


Trang 45

Ghitep;

END.

Bi 4 bi : Thut ton tm ng i theo chiu rng (thut ton duyt theo chiu

rng). Cho th v hng G=(X,E). Hy tm ng i ngn nht gia 2 nh xp v

kt cho trc (ng i ngn nht l xch c 2 u l xp v kt v i qua t cung hoc

t nh nht).

Phn tch bi ton

S dng hng i gii quyt bi ton ny. Cch a mt nh ca th v

ly 1 nh ra s dng phi tun theo mt quy tc nht nh.

S dng mt mng nh du ng i truoc: array[1..Nmax] of integer

Khi i t nh xp m mun c ng i ngn nht n u th ta phi qua nh

truoc [u], mun c ng i ngn nht n truoc [u] th ta phi i qua truoc [truoc

[u] ],

Yu cu: Sau th tc duyt theo chiu rng th ta phi tm c mng truoc.

Khi to gi tr ban u ca mng truoc bng 0.

B1: Khi to hng i.

a xp vo hng i. Gn truoc [xp]: = -1

B2: REPEAT

Ly 1 nh ra khi hng i. Gi s nh u.

Duyt nhng nh k vi u, gi s duyt n nh v.

- Nu trc [v] = 0 th ta a v vo hng i ng thi gn truoc [v]: = u.

Ngha l mun i qua v th phi i qua nh trc l u;

B3: Kim tra v c phi l kt hay khng, nu c th thot khi th tc;

Nu truoc [v] 0 th b qua.

UNITL (hng i rng) or (truoc [kt]0


Trang 46

Hng i rng ngha l khng cn nh no trong hng i.

Trn y l sn ca th tc duyt chiu rng, nu sau nhng bc ny m

truoc [kt] = 0 ngha l khng tn ti ng i t xp n kt.

B4: (Tim ng i thng qua mng truoc)

Dng mt mng duong: array [1..Nmax] of integer biu din cc nh nm

trn ng i. Mng ny th hin ng i ngc t kt v xp. Do khi in kt qu

ra mn hnh ta phi in ngc t cui mng.

Khi u u: = kt;

sold: = 0 (s lng nh trn ng i khi u bng 0)

REPEAT

Tng sold ln 1 n v

gn duong [sold]: = u;

i ngc li bc trc u: = truoc [u];

UNILT u = -1

B5: In kt qu.

Thut ton hi kh hiu. Bn s thy r trong phn ci t chng trnh.

Cu trc d liu

Mng 2 chiu a (ij) biu din ma trn lin thuc cc nh v cc cung ca

th.

Mng truoc v mng duong m t nh trn.

Mng q (quene) m t cu trc ca hng i.

Th tc

Program Duyet _chieu_rong;

Const Nmax = 100;

Type mang = array [1..Nmax] of integer;


Trang 47

Var q,truoc,duong:mang;

A:array [1..Nmax, 1..Nmax] of byte;

Sold; N,xp,kt,qf,ql: integer;

(* qf- queue first l con chy u hng i

ql-queue last l con chy cui hng i*)

Chng trnh minh ha :

Procedure Docfile;

Var ij: integer;

f: text;

Begin

Assign (f, dothi.txt); reset (f);

Readln (f,N);

For i:= 1 to N do

Forj:= 1 to N do

Read (f,a[ i j]);

Close(f);

Readln (f,xp,kt);

End;

(*Th tc khi ng hng i *);

Procedure InitQ;

Begin

ql: = 1

qf: = 0


Trang 48

End;

(*Th tc a mt nh vo cui hng i *)

Prcedure Put (k:integer);

Begin

q[ql]: = k;

inc (ql);

End;

(*Hm ly mt nh ra khi hng i ly u hng i*)

Function Get:integer;

Begin

Get:= q[qf];

inc (qf);

End;

(*Hm kim tra hng i rng hay khng*)

Function Qempty: Boolean;

Begin

Qempty: = (qf>ql);

End;

(*Th tc duyt chiu rng *)

Procedure DuyetCR;


Trang 49

Var v,u:integer;

Begin InitQ;

Put (xp;

Truoc [xp]: = -1;

REPEAT

U: = Get;

(Duyt cc nh k ca u*)

For v: = 1 to N do

If (a[u,v] = 1 and (truoc [v]= 0) then

Beign

(*Nu v k vi u v cha i qua th nh du v a v vo hng i*)

Truoc [v]: = u;

Put (v);

End;

UNILT Qempty or (truoc [kt] 0);

End;

(*Th tc tm mng ng i t xp n kt dng ngc *)

Procedure Timduong;

Var u:integer;

Begin sold: = 0;

u:= kt;

REPEAT


Trang 50

inc (sold);

Duong[sold]:=u;

u:=truoc[u]:

UNILT u = xp;

End;

(*Th tc in kt qu ra mn hnh *)

Procedure Inkq;

Var i:integer;

Begin

If truoc[kt]= 0 then

Writeln (khong co duong di t ,xp, den, kt)

else

Begin Writeln (Duong di t , xp, den, kt);

(*Khi in phi in ngc mng ng t sold tr v 1*)

For i: = sold downto 1 do

Write (duong[i], );

End;

Realn;

End;

BEGIN

Docfile;


Trang 51

Duyet CR;

(*Chc chn rng nu c ng i n nh kt th ta mi tm mng ng *)

If truoc[kt]0 then

Timduong;

Inkq;

END.

d. Bi tp t gii


Trang 52

Bi 7: Cy v cy khung Mc tiu

Mc tiu ca bi ny ngi hc c kh nng:- Xc nh c mt ng i, mt

chu trnh trong th bt k.

- Biu din th trn my tnh bng cc phng php khc nhau.

- p dng c thut ton Kruskal v Prim tm cy khung nh nht ng vi mt

th xc nh. Ci t c thut ton.



a. Nhc li l thuyt

Cc nh ngha: th, cnh, cung, nh k, cy, cy khung.

Cc phng php bin din th trn my tnh

nh l 3.4.2: Cho T l mt th c n 2 nh. Cc iu sau l tng ng:

1) T l mt cy.

2) T lin thng v c n1 cnh.

3) T khng cha chu trnh v c n1 cnh.

4) T lin thng v mi cnh l cu.

5) Gia hai nh phn bit bt k ca T lun c duy nht mt ng i s cp.

6) T khng cha chu trnh nhng khi thm mt cnh mi th c c mt chu

trnh duy nht.

Bi ton cy khung nh nht

Thut ton Kruskal vi tng chn dn

B1: Chn cnh trng s nh nht

B2: Chn cnh trng s nh nht trong cc cnh cn li

B3: Chn cnh trng s nh nht trong cc cnh cn li, m ko to thnh chu

trnh.


Trang 53

Lp li bc 3 cho ti khi chn c n-1 cnh.

Thut ton Prim vi tng lan ta. Gi s lan ta t nh A

B1: Chn cnh trng s nh nht thuc A (tc c 1 u l nh A)

Gi s l cnh (A,B)

B2: Chn cnh trng s nh nht trong cc cnh cn li thuc A hoc B (tc c

1 u l nh A hoc B). Gi s l cnh (C,B) hoc (C,A)

B3: Chn cnh trng s nh nht trong cc cnh cn li thuc A hoc B hoc

C, m ko to thnh chu trnh.

Lp li bc 3 cho ti khi chn c n-1 cnh.

b. bi tp

Bi tp 1: Cho th v hng c trng s sau:

Biu din th dng ma trn k, danh sch cnh(cung), danh sch lien kt.

p dng thut ton Prim( xut pht ti nh 1) hoc Kruskal tm cy khung nh nht

cho th trn.

Bi 2 p dng thut ton Kruskal tm cy khung nh nht ca th cho trong

hnh di y:


Trang 54

c. Hng dn gii

Bi tp 1:

- Thut ton Prim vi tng lan ta, xut pht t nh 1 lan ta ta c cc cnh

to thnh cy khung nh nht ln lt l: (1,5); (5,4);(4,2) v (4,3)

Gi V(E) l tp cc nh(cnh) ca th,

Vt(Et) l tp cc nh(cnh) ca cy khung nh nht cn tm.

Thut ton Prim vi cc bc th t c miu t trong bng sau:

Bc E Et Vt

Khi to {} {1}

1 {(1,5)} {1,5}

2 { (1,5); (5,4); } {1,5,4}

3 { (1,5); (5,4);(4,2) } {1,5,4,2}

4 { (1,5); (5,4);(4,2) v

(4,3) }

{1,5,4,2,3}=V => dng

Ch :

Thut ton Kruskal vi tng chn dn cc nh c trng s nh nht trong cc

cnh cn li, nu ko to thnh chu trnh. Qu trnh lp li cho ti khi chn c n-1

cnh.

Cc cnh c chn to thnh cy khung nh nht ln lt l: (2,4);(4,5);(4,3) v

(1,5)


Trang 55

- Thut ton Prim vi tng lan ta, xut pht t nh 4

lan ta ta c cc cnh to thnh cy khung nh nht ln lt l: (4,2);(4,5);(4,3)

v (1,5)

Vi cc bc th t (t lm)

Bi 2:

Bc khi to. t T:= . Sp xp cc cnh ca th theo th t khng gim ca

di ta c dy:

(3,5) , (4,6) , (4,5) , (5,6) , (3,4) , (1,3) , (2,3) , (2,4) , (1,2)

dy di tng ng ca chng

4, 8, 9, 14, 16, 17, 18, 20, 23.

ba ln gp u tin ta ln lt b sung vo tp T cc cnh (3,5) , (4,6) , (4,5). R

rng nu thm cnh (5,6) vo T th s to thnh 2 cnh (4,5), (4,6) c trong T chu

trnh. Tnh hung tng t cng xy ra i vi cnh (3,4) l cnh tip theo ca dy.

Tip theo ta b sung cnh (1,3), (2,3) vo T v thu c tp T gm 5 cnh:

T = { (3,5) , (4,6) , (4,5) , (1,3) , (2,3) }

Chnh l tp cnh ca cy khung nh nht cn tm.

d. Bi tp t gii

Bi tp 1: Biu din th dng ma trn k, danh sch cnh(cung), danh sch lien

kt.

p dng thut ton Prim v Kruskal tm cy khung nh nht cho th sau:


Trang 56

Bi tp 2 : Mng an ton

Cho mt mng N (N


Trang 57

Ma trn n gi ca mng l mng hai chiu A c N dng v N ct, m gi

tr phn t A[i, j] chnh l n gi t my i sang my j.

Cu 1: Cho trc mt mng, hy kim ra tnh an ton ca mng .

Cu 2: Khi mng khng an ton c php b sung mt s knh truyn

n tr thnh an ton. n gi mi knh truyn b sung theo c coi bng hai ln

gi tr cc i n gi cc knh c. Mi knh b sung c coi c n gi nh

nhau. Hy tm cch b sung cc knh mi m n gi mng l nh nht.

Cu 3: Khi mng an ton hoc sau khi b sung knh mng an ton, hy in

ra n gi mng v ma trn n gi.

D liu vo: cho trong file INP.B2 vi cu trc nh sau:

Dng u tin ghi 2 s n, m cch nhau bi du cch.

Mi dng th i trong s m dng tip theo ghi thng tin v knh ni th i ca

mng gm 3 s d[i], c[i], g[i] trong d[i], c[i] l ch s ca hai my tng ng vi

knh ny v g[i] (nguyn dng) l chi ph truyn mt n v thng tin t my

d[i] n my c[i] theo knh ny. Cc gi tr g[i] cho trc khng vt qu 40 (v

nh vy n gi cc knh b sung khng vt qu 80).

Kt qu: ghi ra file OUT.B2 theo qui cch sau:

Dng u tin ghi 1 s nguyn p th hin mng c an ton hay khng v p c

ngha l s lng knh cn b sung. p=0 c ngha mng an ton; p>0 c ngha mng

khng an ton v cn b sung p knh na mng an ton vi chi ph b sung t

nht.

p dng tip theo ghi p knh b sung. Cch ghi nh trong file d liu vo.

Dng tip theo ghi n gi ca mng an ton.

N dng tip theo ghi ma trn n gi ca mng an ton: mi hng ca ma

trn ghi trn mt dng.

Bi tp 3: Xy dng ng ng nc

C 1 trm cp nc v N im dn c. Hy xy dng chng trnh thit k tuyn

ng ng nc cung cp n mi nh sao cho tng chiu di ng ng phi dng

l t nht. Gi s rng cc ng ng ch c ni gia 2 im dn c hoc gia

trm cp nc vi im dn c.


Trang 58

Bi 8: Tho lun v ci t thut ton tm cy khung nh nht trn th Mc tiu

- Ci t c thut ton xy dng tp cc chu trnh c bn.

- Ci t c thut ton Prim, Kruskal a ra cy khung nh nht ca th cho

trc.

- M rng tng, m rng thut ton Prim.

- Ci t c thut ton xy dng tp cc chu trnh c bn.




a. Nhc li l thuyt

b. bi tp

Bi 1 Tm cy bao trm nh nht ca th G dung thut ton Kruskal

Bi 2 Tm cy bao trm nh nht ca th G dung thut ton Prim

c. Hng dn gii

Bi 1 bi : Tm cy bao trm nh nht ca th G

Phn tch bi ton :

Trc ht ta phi xt v d sau:

8 3

2

1 7 2 3 7

5

1 8


Trang 59

2 1

4 1 7 6

6 1 9

T tng ca thut ton l a tng cnh vo cy bt u t nhng cnh c trng s

nh nht v lm cho n khi cy bao gm tt c cc nh ca th.

Gi s T l cy bao trm cn tm. V tp V l tp bao gm cc tp Gi c tnh cht

nh sau:

- Gi c phn t l cc nh ca th m chng c ni vi nhau bi cc cnh

c a vo T. T mt nh thuc Gi lun c ng i n mt nh khc cng

thuc Gi.

- Giao ca Gi v GJ bng rng.

Vi nh ngha nh trn ta nhn thy r rng rng mi mt tp Gi thuc tp V l

mt cy bao trm con nh nht. Nh vy thut ton ca chng ta nh mt bi ton

quy np l gp nhng cy bao trm ny thnh nhng cy trm ln hn cho n khi

n gm tt c cc nh ca th v ng nhin ta phi ni nhng cy bao trm Gi

ny bng nhng cnh c trng s nh nht. Kt thc khi V = { {1,2,3,n} } = {X};

Ban u:

+ Cy bao trm ca th l rng ngha l cha c cnh no c a vo cy bao

trm. (T = ).

+ Gi chnh l nhng nh ca th => V = { {1}, {2}, {3},{n} }

Thut ton c vit c th vi v d trn nh sau:

- Khi to T = { }

V = { {1}, {2}, {3}, {4}, {5}, {6}, {7}, {8}, {9} }

- a cnh nh nht (1,2) vo cy khi 1 c ni vi 2 nn thay {1} v {2}

trong V bng {1,2}


Trang 60

T = { (1,2) }

V = { {1}, {2}, {3}, {4}, {5}, {6}, {7}, {8}, {9} }

- a cnh (4,5) vo cy:

T = { (1,2); (4,5)}

V = { {1,2}, {3}, {4,5}, {5}, {6}, {7}, {8}, {9} }


T = {(1,2); (4,5); (5,6)}

V = { {1,2}, {3}, {4,5,6}, {7}, {8}, {9} }


T = {(1,2); (4,5); (5,6); (6,9)}

V = { {1,2}, {3}, {4,5,6,9}, {7}, {8} }


T = {(1,2); (4,5); (5,6); (6,9); (1,4)}

V = { {1,2,4,5,6,9}, {3}, {7}, {8} }


T = {(1,2); (4,5); (5,6); (6,9); (1,4); (5,7)}

V = { {1,2,4,5,6,9}, {3}, {8} }


T = {(1,2); (4,5); (5,6); (6,9); (1,4); (5,7); (3,5)}

V = { {1,2,3,4,5,6,7,9}; {8} }


T = {(1,2); (4,5); (5,6); (6,9); (1,4); (5,7); (3,5); (7,8)}

V = { {1,2,3,4,5,6,7,8,9} }

Kt thc v V = {X};


Trang 61

Ch : Th t chn cc cnh a vo cy l u tin cc cnh c trng s b hn.

Cu trc d liu

Ta biu din th di dng lit k cc cung (ch p dng cho th nh)

Nh vy dung mt mng hai chiu:

e: array [1Emax, 1..3] of Integer

vi e [i,3] l trng s ca cung (e[i,1], e[i,2],,emax)

Mng T: array [1Emax, 1..2] of Integer biu din cc cnh c a vo T.

Dng bin solT xc nh s lng cnh a vo cy T;

(khi to solT: = 0 ngha l cy T = )

Biu din V l tp hp ca cc tp hp rt phc tp, do ta dng mt Cu trc d

liu n gin m phng tp V nh sau:

Mng tap V: array[1Nmax] of integer

Trong hai nh i v j thuc cng mt tp Gk no nu nh tap V[i] = tap [j]

Nh vy ban u gi tr ca mng tapV l tapV[i] gn bng i;

mi bc khi ta ghp cc nh ca 2 tp Gk v Gh thnh mt tp th ta ch vic gn

c cc gi tr ca tp Gk bng gi tr ca tp Gh.

V d vi N = 10;

mt bc no

mng tapV = [ 1,2,2,6,6,2,1,4,4,10]

ta hiu l V = { {1,7}; {2,3,6}, {4,5}, {8,9}, {10} }

n bc tip theo ta a cnh (4,6) vo cy ngha l cn gp 2 tp {2,3,6 }v {4,5}

thnh mt. Ta gn tapV[4] = tapV[2;]

tapV[5] = tapV[2;];

Khi mng tapV = [1,2,2,2,2,2,1,4,4,10]

iu ny rt c li khi ta gp hai tp hp con vi nhau.


Trang 62

Vi ton b d liu c m t nh trn ta c phn ci t chng trnh:

Chng trnh minh ha :

Program kruskal;

Const Nmax = 100;

Emax = 1000;

Type mangcanh = array [1..Emax, 1..2] of integer;

Dinh = array [1Nmax] of integer;

Var e,T; Mangcanh;

TapV: Dinh;

N, solE, solT: integer;

(*solE l s cnh ca th cn solT l s cnh c a vo T*)

Prrocudure Doctep;

(* D liu vo gm c:

- Dng u tin cha s N

- T dng th hai tr i mi dng cha 3 s nguyn l 2 nh ca mt cnh v trng

s ca cnh .

*)

Var f: text;

i,j,k: integer;

Begin Assign (f, cung.txt); Reset(f);

Readln (F,N);

SolE: = 0;

While not eof(f) do


Trang 63

begin Read (f,i,j,k);

Inc(solE);

e[solE,1]: = i;

e[solE,2]: = j;

e[solE,3]: = k;

End;

Close (f);

End;

Procedure Sapxep;

(*Dng phng php sp xp ni bt sp xp cc cnh theo th t trng s t

nh ti ln(*)

Var ij : integer;

Beign

For i:= 1 to solE-1 do

For j:= i + 1 to solE do

If e[i,3] > e[j,3] then

Doicho(i,j);

End;

Procedure Doicho (i,j;integer);

Var temp: integer;

Begin

Temp: = e[i,1];


Trang 64

e[i,1]: = e[j,1];

e[j,1]: Temp;

Temp: = e[i,2];

e[i,2]: = e[j,2];

e[j,2]: = Temp;

Temp: = e[i,3];

e[i,3]: = e[j,3];

e[j,3]: = Temp;

End;

Procedure TimCay:

Var i,k, temp:integer;

Begin

solT: = 0; (*T = *)

For i: = 1 to N do

TapV [i]: = i;

K: = 0;

While (not KttapV) and (K


Trang 65

v thc hin tip. Ngc li th b qua cnh ny trnh to thnh chu trnh trong

cy*)

Begin

Inc(solT);

T [solT,1]: = e[k,1];

T [solT,2]: = e[k,2];

(* ng nht gi tr trong 2 tp con cha 2 nh e[k,1] v e[k,2] ca V*). a cnh

vo cy khung.

temp: = tapV [ e[k,2] ];

For i: = 1 to N do

If tapV [i] = temp then

TapV [i]: = TapV [ e[i,2] ];

End;

End;

End;

Function KTtapV: Boolean;

(*Hm kim tra xem cc gi tr ca tp V c bng nhau hay khng*)

Var i: integer;

Begin KTtapV: = True;

For i: = 2 to N do

If tapV [i] tapV [i] then

Begin KTtapV: = False;

Exit;


Trang 66

End;

End;

Procedure GhiTep;

(*Ta ch vic a cy T ra mn hnh*)

Var i: integer;

begin

Writeln ( Cac cung thuoc cay bao trum nho nhat T la:)

For i: = 1 to solT do

Writeln (T [i, l]; ; T[i,2];

Readln;

End;

BEGIN

DocTep;

Sapxep;

TimCay;

GhiTep;

END.

Bi 2 bi : Tm cy bao trm nh nht ca th G

Phn tch bi ton :

Gi s T l cy bao trm ti thiu cn xy dng. Gi x ta bt u t nh u.

Gi U l tp cc nh k vi cc nh c cung ni vi cy T;

Khi to U = u;


Trang 67

T = ;

B1: Chn (u,v) l cnh c trng s b nht vi u

B2: a nh v vo tp U;

B3: Nu U = X th kt thc; ngc li th quay li bc 1;

Cu trc d liu

Ta biu din th di dng lit k cc cung (ch p dng cho th nh)

Nh vy dung mt mng hai chiu:

e: array [1Emax, 1..3] of Integer

vi e [i,3] l trng s ca cung (e[i,1], e[i,2],,emax)

Mng T: array [1Emax, 1..2] of Integer biu din cc cnh c a vo T.

Dng bin solT xc nh s lng cnh a vo cy T;

(khi to solT: = 0 ngha l cy T = )

Biu din V l tp hp ca cc tp hp rt phc tp, do ta dng mt Cu trc d

liu n gin m phng tp V nh sau:

Mng tap V: array[1Nmax] of integer

Trong hai nh i v j thuc cng mt tp Gk no nu nh tap V[i] = tap [j]

Nh vy ban u gi tr ca mng tapV l tapV[i] gn bng i;

mi bc khi ta ghp cc nh ca 2 tp Gk v Gh thnh mt tp th ta ch vic gn

c cc gi tr ca tp Gk bng gi tr ca tp Gh.

V d vi N = 10;

mt bc no

mng tapV = [ 1,2,2,6,6,2,1,4,4,10]

ta hiu l V = { {1,7}; {2,3,6}, {4,5}, {8,9}, {10} }

n bc tip theo ta a cnh (4,6) vo cy ngha l cn gp 2 tp {2,3,6 }v {4,5}

thnh mt. Ta gn tapV[4] = tapV[2;]


Trang 68

tapV[5] = tapV[2;];

Khi mng tapV = [1,2,2,2,2,2,1,4,4,10]

iu ny rt c li khi ta gp hai tp hp con vi nhau.

Tp U l tp cc nh c cung ni vi T, khai bo T,X: set of 1Nmax;

Cc thnh phn d liu khc khai bo ging nh phn trc.

Program Prim;

Const Nmax = 100;

Emax = 1000;

Type mangcanh = array [1Emax, 1...3]of integer;

Tap = set of 1Nmax;

Var e,T: Mangcanh;

VT X: Tap; {X: Tp nh ca th; VT: Tp nh ca cy}

N, solE, solT: integer;

(*Th tc c file v ghi file tng t nh trn*)

Procedure Timcay;

Var k,i,j: integer;

TrongsoMin: Longint;

Begin

solT: = 0; VT = [xp]

X: = [ ];

For i: = 1 to N do

X: = X + [i];


Trang 69

REPEAT

(*Tm cnh (u,v) sao cho u v c trng s nh nht *)

TrongsoMin: = MaxLongint;

For k: = 1 to solE do

Begin

If (e[k,l] in VT and not e[k,2] in VT and (e[k,3] < TrongsoMin)

then

Begin

i: = e[k,1];

j: = e[k,2];

TrongsoMin: = e[k,3;

End;

If not e[k,1] in VT and (e[k,2] in VT and (e[k,3] < TrongsoMin)

then

Begin

i: = e[k,2];

j: = e[k,1];

TrongsoMin: = e[k,3];

End;

(*a j vo tp U*)

vT = VT + [j];

(*a cnh (i,j) vo cy T*)

inc(solT);


Trang 70

T [solT,1]: = i;

T [solT,2]: = j;

UNILT U = X;

End;

BEGIN

Docfile;

Timcay;

Ghifile;

END.

d. Bi tp t gii


Trang 71

Bi 9, 10: Bi ton tm ng i ngn nht Mc tiu

Mc tiu ca bi ny ngi hc c kh nng:- Biu din th trn my tnh bng

cc phng php khc nhau.

- p dng c thut ton ford-bellman v dijkstra tm ng i ngn nht ng vi

mt th xc nh. Ci t c thut ton.



a. Nhc li l thuyt

Thut ton dijkstra vi tng lan ta v lu 1 ch s ti mi nh.

tm li ng i ngn nht, ta lu c tn nh lin k trc n trn ng i

ngn nht.

b. bi tp

Bi tp 1: Biu din th dng ma trn k, danh sch cnh(cung).

p dng thut ton Dijkstra tm ng i ngn nht t nh 5 ti cc nh cn li

cho th sau:

Bi2 Tm ng i ngn nht t 1 n cc nh cn li ca th


Trang 72

Bi 3 a th

Nh bc a th c vic bn, bc a th rt st rut mun v nh ngay

nhng bc phi a mt th quan trng cho anh B qun Ba nh. Hin ti

bc qun Thanh Xun. ng i t ch bc ti nh anh B phi i qua mt s

qun huyn khc vi khong cch gia cc qun nh sau. Hy gip bc a

th tm ng i ngn nht vi thi gian t nht bc c th v nhanh chng

gii quyt cng vic gia nh.

Bi 4 H bt mi

Mt con h ng trn i cao nhn xung khu rng v pht hin ra 5 con mi

5 v tr khc nhau. Con h mun bt con Nai trong s 5 con vt l Nai, G,

Hu, Th, B n tht. Khi con H mun bt con Nai , n phi chy qua

ng m 4 con vt cn li ang ng. V tr v khong cch m cc con vt

c biu din nh sau. Tm ng i qung ng m con H phi chy

l ngn nht bt c con Nai.

Bi 5 Mo bt chut

Mo Tom(1) rnh ch chut nh Jeny(4) ang n trm phomat. T ch Tom n

Jeny c ci x nc(2), chu cy cnh (3), gh sofa(5) vi thi gian Tom chy

n cc vt trn v gia cc vt cng nh thi gian Tom chy t cc vt

n Jeny nh sau: x nc chu cy cnh


Trang 73

Tom Jeny

gh sofa

Tom c th chy n np ch cc vt trn rnh v bt Jeny. Tm thi gian

nh nht Tom chy n bt Jeny m khng b Jeny pht hin(nh np vo cc

vt c trn ng i)

Bi 6 Cho th c trng s G = (X,E). Tm ng i ngn nht gia 2 nh xp v

kt thc ca th

c. Hng dn gii

Bi 1:

Biu din th dng ma trn k nh sau:

0 18 M 13 12

18 0 11 3 M

M 11 0 5 9

13 3 5 0 4

12 M 9 4 0

Dng thut ton Dijkstra gim trng s 1 ln ti mi nh ta c:

K hiu:


Trang 74

D[i] : Ch s ti nh i (Tng trng s t nh xut pht ti nh i, c cp nht

theo tng bc ca thut ton)

T[i] : Tn nh k ngay trc nh i trn ng i ngn nht c cp nht theo

tng bc ca thut ton

Bc nh 5

c

chn

Cc nh

cha xt

D[1],T[1] D[2],T[2] D[3],T[3] D[4],T[4]

Khi to

1

2

3

4

5

4

2

3

1

1,3,4,2

1,3,2

1,3

1

12,5

12.5

12,5

12,5

12,5

7,4

7,4

7,4

7,4

9,5

9,5

9,5

9,5

9,5

4,5

4,5

4,5

4,5

4,5

Trong , bc khi to c cho trong dng u tin.

Thc hin bc 1, trong D[v], vi v=1, 2, 3, 4 dng u tin, th D[4 ]

= 4 l nh nht nn nh 4 c chn. Ta xc nh li cc D[v] cho cc nh cn li

1, 2 v 3. Ch c D[2] l nh i v D[2] = 7, v D[4] + c(4,2) = 4 + 3 = 7 < .

Tip tc thc hin cc bc 2, 3, 4 ta thu c di ng i ngn nht t

nh 0 ti cc nh 1, 2, 3, 4 c cho dng cui cng trong bng, chng hn

di ng i ngn nht t 5 ti 2 l D[2] = 7 v l di ca ng i 542.

Bi2 Tm ng i ngn nht t 1 n cc nh cn li ca th


Trang 75

Kt qu tnh ton theo thut ton c trnh by theo bng di y: Quy c vit

hai thnh phn ca nhn theo th t d[v]. nh c nh du * l nh c chn

c nh nhn bc lp ang xt, nhn ca n khng bin i cc bc tip theo, v

th ta nh du (-).

Bc lp nh 1 nh 2 nh 3 nh 4 nh 5 nh 6

Khi 0, 1 1, 1* , 1 , 1 , 1 , 1

1 - - 6, 2 3, 2* , 1 8, 2

2 - - 4, 4* - 7, 4 8, 2

3 - - - 7, 4 5, 3*

4 - - - 6, 6* -

Bi 3 Nh bc a th c vic bn, bc a th rt st rut mun v nh ngay

nhng bc phi a mt th quan trng cho anh B qun Ba nh. Hin ti bc

qun Thanh Xun. ng i t ch bc ti nh anh B phi i qua mt s qun

huyn khc vi khong cch gia cc qun nh sau


Trang 76

Hy gip bc a th tm ng i ngn nht vi thi gian t nht bc c th v

nhanh chng gii quyt cng vic gia nh.

*Phn tch bi ton: Bc a th cn chuyn th t qun Thanh Xun ti Qun Ba

nh nhng li khng c ng i trc tip t qun Thanh Xun ti qun Ba nh

m phi i qua mt s qun khc. Vi im xut pht l qun Thanh Xun tm cc

con ng c th i ti qun Ba nh v tnh khong cch gia hai qun khi i

qua cc con ng khc nhau.

Ta tnh khong cch t Qun Thanh Xun (Q.TX) ti cc Q.Thanh Tr (Q.TT),

Q.Hong Mai (Q.HM), Q.Ty H (Q.TH), Q.H ng (Q.H). Ta thy khng c

ng i trc tip t Q.Thanh Xun n Q.H ng, nn cn 3 s la chn l i

qua Q.TT, Q.HM v Q.TH. Ta c khong cch tng ng l:

Q.TX Q.TT = 1km;

Q.TX Q.HM = 7km;

Q.TX Q.TH = 6km;

NX: Qung ng t Q.TX n Q.TT ngn nht vi 1km. Vy ta s chn v nh

du li i ny.

Tip tc tnh ng i t Q.TT n Q.HM, Q.TH, Q.H v Q.B . Ta c:

Q.TT Q.HM = ;

Q.TT Q.TH = 3km;

Q.TT Q.H = 4km;


Trang 77

Q.TT Q.B = 1km;

NX: Qung ng t Q.TT n Q.B l ngn nht 1km v l ch ca chuyn i.

Vy tng khong cch t Q.TX Q.B l 1+1=2km.

Gi s cc nh 1,2,3,4,5,6 tng ng vi cc qun Thanh Xun, Hong Mai, Ty

H, Thanh Tr, H ng v qun Ba nh. Ta c th sau:

Tm ng i ngn nht t qun Thanh Xun ti Ba nh l tm ng i ngn

nht gia t nh 1 ti nh 6 ca th trn. p dng thut ton Dijkstra ta c bng

sau:

1 2 3 4 5 6

Khi to -1,0* 1, 1, 1, 1, 1,

1 ---- 1,7 1,6 1,1* ---- ----

2 ---- ---- 4,4 ---- 4,5 4,2*

3 ---- ---- ----* ---- 6,4 ----

4 ---- ---- ---- ---- ----* ----

T bng trn ta thy con ng ngn nht i t nh 1 ti nh 6 l 1 4 6

vi trng s l 2, ni cch khc bc a th c th i t qun Thanh Xun ti qun

Thanh Tr v ti qun Ba nh vi qung ng l 2km.


Trang 78

Bi 4 Mt con h ng trn i cao nhn xung khu rng v pht hin ra 5 con mi

5 v tr khc nhau. Con h mun bt con Nai trong s 5 con vt l Nai, G,

Hu, Th, B n tht. Khi con H mun bt con Nai , n phi chy qua

ng m 4 con vt cn li ang ng. V tr v khong cch m cc con vt c

biu din nh sau. Tm ng i qung ng m con H phi chy l ngn nht

bt c con Nai.

*Phn tch bi ton: Con h khng th bt c tt c 5 con vt , v n ch c th

chn 1. Trong cc con ng i n ch con Nai m n mun bt , c nhng

con ng di v ngn khc nhau. N phi tnh ton sao cho qung ng chy n

con Nai l ngn nht phng con vt nhn thy t xa v b chy.

Gi v tr H, B, Th, Hu, Nai tng ng l 1,2,3,4 vi cc nh ca th sau:

i t 1 n 5 ta tnh khong cch t 1 n cc nh cn li:

1 2 = 4

1 3 =


Trang 79

1 4 =

1 5 =

Ta thy c ng i duy nht l 12.

Tip tc tnh khong cch t 2 3, 4, 5 ta c:

2 3 = 10

2 4 = 2

2 5 = 8

NX: Qung ng 2 4 l nh nht. nh du v chn con ng ny.

Khi chn ng i, nu ng no c chn v nh du s khng c xt li

na. Vy cn li 2 nh cha xt l 3 v 5, nhng khng c ng i trc tip t 4

3, ch cn ng 4 5 vi khong cch l 1km v l mc ch ca bi ton.

KL: ng i ngn nht t 1 5 l 1 2 4 5 vi khong cch l 1+4+2=7

p dng thut ton Djstral ta c bng sau:

1 2 3 4 5

Khi to -1,0* 1, 1, 1, 1,

1 ---- 1,4* ---- ---- ----

2 ---- ---- 2,14 2,6* 2,12

3 ---- ---- ---- ---- 4,7*

4 ---- ---- ---- ---- ----

Nhn vo bng trn ta thy ng i ngn nht t 1 n 5 l 1 - 2 - 4 - 5 vi trng s

l 7.

Ni cch khc con H mun bt v n tht con Nai th H phi i qua ch con B v

con Hu.

Bi 5


Trang 80

Mo Tom(1) rnh ch chut nh Jeny(4) ang n trm phomat. T ch Tom n

Jeny c ci x nc(2), chu cy cnh (3), gh sofa(5) vi thi gian Tom chy

n cc vt trn v gia cc vt cng nh thi gian Tom chy t cc vt

n Jeny nh sau: x nc chu cy cnh

Tom Jeny

gh sofa

Tom c th chy n np ch cc vt trn rnh v bt Jeny. Tm thi gian

nh nht Tom chy n bt Jeny m khng b Jeny pht hin(nh np vo cc

vt c trn ng i)

Bi lm:

Gi s v tr cc i tng tng ng vi cc nh 1, 2, 3, 4, 5 ca th sau:

Phn tch tng t nh 2 bi ton trn ta s tm c con ng cn tm

p dng thut ton Dijkstra tm ng i ngn nht t 1 4

1 2 3 4 5

Khi to -1,0* 1, 1, 1, 1,

1 ---- 1,3* ---- ---- 1,12


Trang 81

2 ---- ---- 2,8 ---- 2,7*

3 ---- ---- ---- 5,15* ----

4 ---- ---- ---- ---- ----

Da vo bng trn ta c con ng ngn nht l 1 2 5 - 4 vi thi gian l 15

Bi 6 Cho th c trng s G = (X,E). Tm ng i ngn nht gia 2 nh xp v

kt thc ca th.

Phn tch bi ton :

Phng php 1: Ti mi nh thc hin vic gim trng s nhiu ln.

Bi ton ny cng dng mng truoc nh du cc nh trn ng i nh trn v

dng cu trc hng i.

C mt im mi l n cn thm mt mng trongso: array[1Nmax] of (tu vo bi ton c th l kiu real hoc kiu longint).

Trongso[i] bng tng trng s ca cc cung nm trn ng i tnh t nh xp cho

n i. Bi ton ca chng ta tr thnh tm mng trongso[i] sao cho n t gi tr

nh nht.

Thut ton ca phn ny rt ging thut ton duyt chiu rng nhng ci tin mt

cht l c gn thm mng trng s.

Khi duyt cc nh k ca mt nh u va ly ra khi hnh i, gp mt nh i k

vi u ta khng cn bit truoc[i] l nh no, m ta ch quan tm ng i t u n i

c ngn hn ng c khng (trongso[i]???trongso [u] +a[u,i].

Trongso [i] chnh l ng i c n nh i, trongso [u] + a[u,i] l ng i mi t

nh u i n.

Nu ng mi ngn hn ngha l trongso[u] + a[u,i]


Trang 82

truoc[i]:= u;

Thut ton c trnh by c th nh sau:

Khi to:

- Mng trongso [i] = Maxlongint;

- Truoc [xp]: = -1;

- Khi ng hng i v a xp vo;

Thc hin bc lp.

REPEAT

- Ly 1 nh khi hng i gn vo bin u;

- Duyt cc nh i k vi u;

(Nu ng i mi ngn hn ng i c th thc hin gim trng s cho i)

Nu trongso[i]>trongso[u] +a[u,i] th

+ trongso[i]:= trongso[u] + a[u,i];

+ truoc [i]:= u;

UNILT ;

Chng trnh minh ha :

Program DIJSTRA_phuongphap1;

Const Nmax = 100;

Type MATRANKE = array[1Nmax, 1Nmax] of integer;

MANG = array [1Nmax] of Longint;

Var a: MATRANKE;

q, duong, truoc, trongso: MANG;


Trang 83

N, sold, xp, kt, qf, ql: Integer;

fifo: text:

(*fi: file input la tep doc du lieu vao

fo: file output la tep doc du lieu ra*)

Procedore Doctep;

Var i,j: integer;

Begin

Asign (fi,matran:inp); Reset(fi);

Readln (fi, N);

For i: = 1 to N do

For j: = 1 to N do

Read (fi, a[i,j]);

Readln (f,xp,kt);

Close (f);

End;

Procedure InitQ:

(*Thu tuc khoi dong hang doi *)

Begin

qf: = 0;

ql:= 1;

End;


Trang 84

Procedure Put (u:integer);

(*Dua mot phan tu vao hang doi *)

Begin

Inc(ql);

q[ql]: = u;

End;

Function Get: Integer;

(*Ham lay mot phan tu ra khoi hang doi*)

Begin

Get:= q[qf];

inc [qf];

End;

Function Qempty: Boolean;

(*Ham kiem tra hang doi da rong hay cha*)

Begin

Qempty:= (qf>ql);

End;

Procudure Dijstra

(*Thu tuc duyet Dijstra theo phuong phap 1: giam trong so nhieu lan*)

Var i, u: integer;


Trang 85

Begin (*cac buoc khoi tao (giong trong phan thuat toan)*)

truoc[xp]:= -1;

Fullhar() = For i:= 0 to N do Trongso[i]:= MaxLongint;

Trongso[xp]:= 0;

InitQ;

Put(xp);

(*Bat dau vong lap*)

REPEAT

(*Lay mot dinh khoi hang doi gan vao bien u*)

u:= Get;

For i:= 1 to N do

(*Duyet tat ca cac dinh i cua do thi*)

If (a[u,i]>=0) and (trongso [i]>trongso [u] + a[u,i]then

(*Neu dinh i ke voi u va duong di cu lon hon duong di moi

thi:*)

Begin (*Gan lai gia tri duong di moi*)

trongso [i]:= trongso [u] +a[u,i];

(*danh dau duong di moi den i qua dinh truoc do la u*)

truoc [i]:= u;

Put(i);

End;

UNILTQempty: (*Ket thuc khi khong di duoc nua*)

End;


Trang 86

Procedure Timnguoc;

Var u:integer;

Begin

sold:= 0; (*Mng ng rng*)

(*Ban u gn u bng kt*)

u:= kt;

Repeat

(*Vi mi bc lp ghi u vo mng duong*)

Inc(sold);

duong[sold]:= u;

(*Mun n bc tip theo phi gn u = truoc [u]*)

u:= truoc[u];

Unilt u = -1; (*Ket thuc khi u:= -1 do luc truoc ta gan truoc [xp]= -1).

End;

procedure Ghitep:

Var ij:integer;

Begin

Asign (fo, kq.out); Rewrite (fo);

If truoc [kt] 0 then

Begin Timnguoc;

For i:= sold downto 1 do

Write (fo, duong [i], )


Trang 87

End else

Writeln (f,0);

Close (fo);

End;

BEGIN

Doctep;

Dijstra;

Ghitep;

END.

Phng php 2: Gim trng s 1 ln ti mi nh.

lm r thut ton ny ta xt v d c th sau:

G= (X,E) a ra hnh v

Mng ma trn trng s nh sau:

3

5

1 4

2 6

7

Ta tm ng i bt u t nh 1.

-1 1 2 5 -1 -1 -1

1 -1 -1 3 -1 -1 -1

2 -1 -1 7 -1 -1 -1

5 3 7 -1 5 -1 13

-1 -1 -1 5 -1 2 -1

-1 -1 -1 -1 2 -1 6

-1 -1 -1 13 -1 6 -1


Trang 88

Cc bc thc hin ca thut ton i vi v d ny nh sau:

(M = MaxLongint). Trnh t thc hin mi bc t tri sang phi.

Bc

nh c

trng s

nh nht

trong B

Tp A Tp B Mng trongso Mng Truoc

B1 [1] [2,3,4,5,6,7] [0,M,M,M,M,M,M] [-1,0,0,0,0,0,0,]

B2 [1] [2,3,4,5,6,7] [0,1,2,5,M,M,M] [-1,1,1,1,0]

B3 2 [1,2] [3,4,5,6,7] [0,1,2,4,M,M,M] [-1,1,1,2,0,0,0]

B4 3 [1,2,3] [4,5,6,7] [0,1,2,4,M,M,M] [-1,1,1,2,0,0,0]

B5 4 [1,2,3,4] [5,6,7] [0,1,2,4,8,M,17] [-1,1,1,2,4,0,4]

B6 5 [1,2,3,4,5] [6,7] [0,1,2,4,8,10,17] [-1,1,1,2,4,5,4]

B7 6 [1,2,3,4,5,6] [7] [0,1,2,4,8,10,16] [-1,1,1,2,4,5,6]

B8 7 [1,2,3,4,5,6,7] [] nt Nt

B9 Kt thc

Bi ton ny cng dng mng trc nh du cc nh trn ng i nh trn

nhng khng dng cu trc hng i.

Dng mt mng trongso: array [1Nmax] of (tu vo bi ton

c th l kiu real hoc kiu longint).

Trongso [i] bng tng trng s ca cc cung nm trn ng i t nh xp cho n

i. Bi ton ca chng ta tr thnh tm mng trongso [i] sao cho n t gi tr nh

nht.

Chia tp cc nh ca th X thnh 2 phn

a. Tp cc nh c trng s c gn c nh A.

b. Tp cc nh c trng s c gn tm thi B.


Trang 89

Trong sut qu trnh duyt nhng nh c a vo tp A th trng s ca nh

khng thay i v l di ng i ngn nht t xp n n, nhng nh tp B c

trng s thay i n khi nh nht th c a vo tp A. Cch thay i trng s

nh nht c th hin nh sau:

B1: Khi to

Khi to cc gi tr ca mng trongso = Maxlongint (hoc Maxreal)

Lp hai tp A v B. Ban u A rng cn B = X

Duyt tt c cc nh u k vi nh xp th gn trongso [u] = a[u, xp] v truoc[u] =

xp;

REPEAT

B2: Tm nh v thuc tp B c trng s nh nht.

a nh v vo tp A (B/{v}; A + {v})

B3: Duyt tt c cc nh i thuc B m k vi v

Nu trongso[i]>trongso[v]+ a[i,v] th

+ Gn li trongso [i]:= trongso[v] + a[i,v];

+ ng thi gn truoc [i]:= v;

UNILT ;

Cu trc d liu

Ma trn k biu din quan h ca th:

a(i;j) = -1 nu i khng ni vi j

>= nu i ni vi j v a(a,j) l trng s ca cung 9i,j).

Mng truoc v trongso c m t nh trn.

Mng duong lu cc nh trn ng i.

biu din tp A v B ta dng kiu tp hp.

tap:set of


Trang 90

Trong phn ci t trn, khai bo tp A ch l minh ho cho thut ton cn trong bi

ton ny khng cn thit phi s dng tp A, ta hiu ngm khi loi mt phn t khi

tp B th ngha l ta a n vo tp A.

Chng trnh minh ha :

Program Difstra_Phuongphap2;

Const Nmax = 100;

Type Mang = array [0Nmax] of Longint;

Tapcacdinh = set of aNmax;

Var a: array[1Nmax, 1Nmax] of integer;

duong, truoc, trongso: Mang;

N, sold, xp, kt: Integer;

tap A, tap B: set of Tapcacdinh;

(*Ch trnh by th tc Dijstra, cc th tc khc tng t nh phn trn *)

Prcedure Dijstra;

Var i, u, v: integer;

Begin (*B1 khi to cc gi tr cn thit *)

For i;= 0 to N do

Trongso [i]:= MaxLongint;

(*Khi to tp B l tp tt c cc nh ca th*)

tapB: = [ ];

For i:= 1 to N do

tap B:= tapB +[i];

(*a xp vo tp A*)

tapA:= [xp]


Trang 91

tapB: = tapB - [xp];

truoc[xp]:= -1;

(*Tm cc nh k vi xp gn trng s v nhn cho nhng nh *)

For i:= 1 to N do

if a [xp;i]>=0 then

Begin

trongso[i]:= a[xp,i];

truoc[i]:= xp

End;

REPEAT

(*Tm nh v c trng s nh nht trong tp B v a vo tp A, loi khi tp

B*).

j: = Dinh_co_trong_so_min;

tapA:= tapA + [j];

tapB:= tapB + [j];

(*Tm tt c cc nh thuc B k vi nh v v thc hin vic gim trng s

cho chng*).

For i:= 1 to N do

If (i in tapB) and (a [i,j]>=0 and(trongso[i]>trongso[j] + a[i,j]) then

Begin

trongso [i]: = trongso[j] + a[i,j];

truoc[i]:= j;

End;

UNILT tapB = [ ];


Trang 92

End;

(*Hm tm nh c trng s nh nht trong cc nh thuc tp B*)

Function Dinh_co_trong_so_min: Integer;

Var i,t: integer;

Begin tS [it]:= maxint

For i:= 1 to N do

If (i in tap B and (trongso[i] < trongso[t] then

t:=i;

Dinh_co_trong_so_+min:= t

End;

d. Bi tp t gii

Bi tp 1: p dng thut ton Dijkstra tm ng i ngn nht cho th sau:


Trang 93

Bi tp 2: Di chuyn trn cc hnh trn

Cho N hnh trn (nh s t 1 n N). Mt ngi mun i t hnh trn ny

sang hnh trn khc cn tun theo qui c:

- Nu khong cch gia 2 im gn nht ca 2 hnh trn khng qu 50 cm th

c th bc sang.

- Nu khong cch ny hn 50cm v khng qu 80cm th c th nhy sang.

- Cc trng hp khc khng th sang c.

Mt ng i t hnh trn ny sang hnh trn khc uc gi l cng "tt" nu

s ln phi nhy l cng t. Hai ng i c s ln nhy bng nhau th ng i no

c s hnh trn i qua t hn th ng i "tt" hn.

Cc hnh trn c cho trong mt file vn bn, trong dng th i m t

hnh trn s hiu i (i = 1, 2,..., N) bao gm 3 s thc: honh tm, tung tm,

ln bn knh (n v o bng mt).

Lp trnh c cc hnh trn t mt file vn bn (tn file vo t bn phm), sau

c mi ln c s hiu hnh trn xut pht S v hnh trn kt thc T t bn phm,

chng trnh s a ra ng i t S n T l "tt nht" theo ngha nu (hoc

thng bo l khng c).

Yu cu ng i c vit di dng mt dy cc s hiu hnh trn ln lt

cn c i qua trong ni r tng s cc bc nhy, tng s cc hnh trn i qua

v nhng bc no cn phi nhy.

Gii hn s hnh trn khng qu 100.

Bi tp 3: Tm hnh trnh tn t xng nht


Trang 94

Trn mt mng li giao thng, mt ngi mun i t im A n im B

bng xe my. Xe cha c ti a 3 lt xng v chy 100km ht 2,5 lt. Cc trm

xng ch c t cc im dn c, khng t gia ng v ngi ny khng

mang theo bt k thng cha xng no khc. Hy vit chng trnh nhp vo mng

li giao thng v xc nh gip ngi ny tuyn ng i t A n B sao cho t

tn xng nht.

Bi tp 4: Di chuyn gia cc o

Trn mt o quc, c N hn o. Gi s tt c cc o u c hnh dng l

hnh ch nht nm ngang. Trn mi hn o c th c sn bay nm trung tm o,

c th c cng nm 4 gc o. Trn mi o u c tuyn ng xe but ni 4

gc o vi nhau v vi trung tm o. Gia 2 o c th i li bng my bay nu

c 2 o u c sn bay v c th i li bng tu nu c 2 o u c cng.

Gi s rng:

- Cc tuyn ng (b, khng, thy) u l ng thng.

- Chi ph cho mi km v tc ca mi loi phng tin l:

Phng tin Tc

(km/h)

Chi ph

(/km)

My bay 1000 1000

Xe but 70 100

Tu thy 30 50

Hy vit chng trnh xc nh tuyn ng v cch di chuyn gia 2 hn

o trong o quc sao cho:

- Thi gian di chuyn t nht.

- Chi ph di chuyn t nht.

- Thi gian di chuyn t nht nhng vi mt s tin chi ph khng qu

ng.


Trang 95

- Chi ph di chuyn t nht nhng vi thi gian di chuyn khng vt qu T

gi.

Bi tp 5: Tm ng ngn nht

Gi s X l tp cc khu dn c, U l tp cc ng s ni lin cc khu . Ta

gi s mi ch giao nhau ca cc con ng u thuc X. Vi con ng u, s l(u)

l di ca u tnh bng km. Hy ch ra tuyn ng i t mt khu i sang khu j sao

cho tng chiu di l nh nht.

Bi tp 6: ng i trn li

Cho 1 ma trn A[M, N], mi phn t ca n cha 1 s t nhin. T 1 (i, j)

ta c th i sang k n (c chung 1 cnh) nu gi tr ca k ny nh hn gi tr

lu trong (i, j). Hy tm 1 ng i t (i, j) ti (k, l) trn ma trn sao cho phi i

qua t nht. Hy tm 1 ng i t (i, j) ti (k, l) trn ma trn sao cho tng gi

tr cc phi i qua nh nht.

Bi tp 7 : Tm ng vi chi ph phi tr cho php

C N thnh ph c nh s t 1..N ni vi nhau bng cc on ng mt

chiu. Mi on ng bao gm 2 thng s : di v chi ph i ca on ng.

A sng ti thnh ph 1 v A mun di chuyn n thnh ph N nhanh nht c

th.

Bn hy gip A tm ra ng i ngn nht t thnh ph 1 n thnh ph N

m A c kh nng chi tr tin.

D liu vo : ROADS.IN

- Dng u tin cha s nguyn K, 0


Trang 96

+ L l di ca on ng, 1


Trang 97

Bi 12: Bi ton lung cc i trong mng Mc tiu

- Trnh by c t tng ch o ca thut ton Ford_Fulkerson v ci t c

thut ton.



a. Nhc li l thuyt

nh ngha 1. Ta gi mng l th c hng G = (V,E), trong c duy

nht mt nh s khng c cung i vo gi l im pht, duy nht mt nh t khng

c cung i ra gi l im thu v mi cung e = (v,w) E c gn vi mt s

khng m c(e) = c(v,w) gi l kh nng thng qua ca cung e.

nh ngha 2. Gi s cho mng G = (V,E). Ta gi lung f trong mng G =

(V,E) l nh x f: E R+ gn cho mi cung e =(v,w)

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