Gtth Lap Trinh Nang Cao_dhcl

8/6/2019 Gtth Lap Trinh Nang Cao_dhcl

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LI M U

Gio trnh thc hnh ny c vit theo gio trnh Lp trnh nng cao nhm mc ch

lm ti liu cho sinh vin nm th 2 thc hnh mn hc ny. Ni dung ca gio trnh

gm 4 chng th hin c bn cc k thut lp trnh thng gp i vi sinh vin.

Chng 1. K thut lp trnh quy. Chng 2. Sp xp. Chng 3. i s ma trn. Chng 4. Mt s thut gii trn th.

Chng 1 th hin mt s k thut lp trnh lm nn tng cho cc chng sau. i vi

quy phi tuyn ch yu ta s dng k thut tm kim theo chiu su. K thut ny

c p dng trong chng 4 tm ng i trn th. Tuy nhin, y ta chatrnh bi k thut duyt theo chiu su bng cch kh quy. K thut ny s c

trnh bi trong gio trnh L thuyt th v thut gii.

Chng 2 th hin mt s thut ton sp xp nhm gip sinh vin so snh v nh gi

thut ton sp xp no s tt hn.

Chng 3 th hin phng php gii h phng tr nh tuyn tnh bng phng php

phn r ma trn bng thut ton Crout.

Chng 4 th hin mt s thut gii tm ng i c bn trn th p dng k thut

nh du nh, nh du cnh v k thut tham n.

V thi gian phn b ging dy theo chng trnh khung v ni dung ca mn hc ny

nn gio trnh khng trnh khi nhng khim khuyt. Rt mong nhn c s gp

ca tt c cc bn quan tm n gio trnh ny.

Ngy 24 thng 04 nm 2010

Tc gi


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MC LC

CHNG 1. K THUT LP TRNH QUY 1Bi tp 1. Tm phn t Fibonacci th n 1Bi tp 2. Tnh X ly tha n 1Bi tp 3. Thut ton Euclide tm c chung ln nht 2

Bi tp 4. Tm

c chung l

n nh

t c

a n s

nguyn

3Bi tp 5. Tnh n giai tha 4

Bi tp 6. T hp chp k ca n phn t 4Bi tp 7. Tnh tng n phn t trong danh sch 5Bi tp 8. quy h tng 6Bi tp 9. Tch n phn t trong danh sch 7Bi tp 10. m s ln xut hin ca phn t x trong danh sch 8Bi tp 11. Thp H Ni 9Bi tp 12. Lit k tt c dy nh phn di k 10Bi tp 13. Chnh hp khng lp chp k ca n phn t 12Bi tp 14. Hon v mng s nguyn c n phn t 14Bi tp 15. t n qun hu trn bn cvua 16

Bi tp 16. Mi tun 18

CHNG 2. SP XP 20Bi tp 1. Thut ton Bubble Sort 20Bi tp 2. Thut ton Selection Sort 23Bi tp 3. Thut ton Insertion Sort 26Bi tp 4. Thut ton Quick Sort 28Bi tp 5. Thut ton Heap Sort 30Bi tp 6. Thut ton Merge Sort 34

CHNG 3. I S MA TRN 36

Bi tp 1. Nhp xut ma trn 36Bi tp 2. Mt s php ton trn ma trn 37Bi tp 3. H phng trnh tuyn tnh dng tam gic trn 39Bi tp 4. H phng trnh tuyn tnh dng tam gic di 41Bi tp 5. Thut ton phn r ma trn A = LU 44Bi tp 6. Gii h phng trnh tuyn tnh da vo phn r LU 46Bi tp 7. nh thc ca ma trn 49

CHNG 4. MT S THUT GII TRN TH 51Bi tp 1. Xt tnh lin thng ca th 51Bi tp 2. m s thnh phn lin thng 53Bi tp 3. Tm mi ng i t gia hai nh 56Bi tp 4. ng i Hamilton 58Bi tp 5.ng i Euler 61Bi tp 6. Thut ton Dijkstra tm ng i ngn nht 63Bi tp 7. Thut ton Prim tm cy bao trm ti tiu 65Bi tp 8. Thut ton Kruskal tm cy bao trm ti tiu 67

TI LIU THAM KHO 70


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CHNG 1. K THUT LP TRNH QUY

Bi tp 1. Tm phn t Fibonacci th n

Vit chng trnh tm phn t Fibonacci th n c nh ngha quy nh sau:

1,21

10,1)(

nnFnF

nnnF

Ci t:#include

#include

/*Ham tra ve so nguyen tinh gia tri Fibonacci thu n*/

int F(int n) {

if(n==0 || n==1)

return 1;else

return F(n-1) + F(n-2);

}

/*Chuong trinh chinh*/

void main() {

clrscr();

int n;

coutn;

cout


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Bi tp 4. Tm c chung ln nht ca n s nguyn

Vit chng trnh tm c chung ln nht ca n s nguyn dng 10 ,..., naa c nh

ngha quy nh sau:

1,1,,...,,

1,,,...,

201

0

10

nnaaUCaUCLN

nanaaUC

nn

n

Ci t:

#include #include

/*Ham tra ve uoc chung lon nhat cua a va b*/

int UCLN(int a, int b) {

if(a==b)

return a;

else if(a>b)

return UCLN(a-b,b);

else

return UCLN(a,b-a);

}

/*Ham tra ve uoc chung lon nhat cua n phan tu duoc luu tru trong mang 1 chieu a*/

int UC(int a[], int n) {

if(n==1)

return a[0];

else

return UCLN(a[n-1],UC(a,n-1));

}

void main() {

clrscr();

int *a,n;

coutn;

a = new int[n];

cout


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Bi tp 5. Tnh n giai tha

Vit chng trnh tnh n! c nh ngha quy nh sau:

1,!1*

0,1!

nnn

nn

Ci t:#include

#include

/*Ham tra ve so nguyen tinh n! (Factorial)*/

long int Fac(int n) {

if(n==0)

return 1;

else

return n*Fac(n-1);

}


void main() {

clrscr();

int n;

coutn;

cout


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Bi tp 8. quy h tng

Vit chng trnh tnh nX v nY c xc nh nh sau:

11

11

0

0

2

1

1

nnn

nnn

YXY

YXX

Y

X

Ci t:#include

#include

long int Y(int n);

long int X(int n) {

if(n==0)

return 1;

else

return X(n-1) + Y(n-1);

}

long int Y(int n) {

if(n==0)

return 1;

else

return 2*X(n-1)*Y(n-1);

}


void main() {

clrscr();

int n;

coutn;

cout


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Bi tp 9. Tch n phn t trong danh sch

Vit chng trnh tnh tch n phn t 10 ,..., naa c nh ngha quy nh sau:

1,1,,...,

1,,,...,

201

0

10 nnaaSa

nanaaS

nnn

Ci t:#include

#include

/*Ham tra ve so nguyen tinh tich n phan tu trong mang a*/

long int S(int a[], int n) {

if(n==1)

return a[0];

else

return a[n-1]*S(a,n-1);

}/*Chuong trinh chinh*/

void main() {

clrscr();

int *a,n;

coutn;

a = new int[n];

cout


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Bi tp 10. m s ln xut hin ca phn t x trong danh sch

Vit chng trnh m s ln xut hin ca s nguyn x trong danh sch

10 ,..., naaA vi n phn t. Thut ton quy c th hin nh sau:

xanxnAFind

xanxnAFind

n

xnAFind

n

n

0,,1,

0,,1,1

0,0

,,

Ci t:#include

#include

/*Ham tra ve so lan xuat hien cua x trong danh sach A*/

int Find(int a[], int n, int x) {

if(n==0)

return 0;

else

if(a[n-1]==x)

return 1+Find(a,n-1,x);

else

return Find(a,n-1,x);

}


void main() {

clrscr();

int *a,n,x;

coutn;

a = new int[n];

cout


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Bi tp 1 1 . Thp H Ni

M t bi ton: chuyn n a t ct 1 sang ct 2 ly ct 3 lm trung gian. Th t cca c sp xp t nh n ln (a ln nm pha di).

V d: n = 3 ta c cc bc chuyn 12, 13, 23, 12, 31, 32, 12.

Ci t:#include "conio.h"

#include "iostream.h"

/*Thap Ha Noi*/

void Move(int n, int a, int b) {

if (n==1)

cout


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Bi tp 12. Lit k tt c dy nh phn di k

Chnh hp lp chp k ca n phn t l mt nhm c th t gm k phn t ly t nphn t cho, trong mi phn t c th c mt 1, 2, , k ln trong nhm to

thnh.

Phng php: ta lit k tt c chnh hp c lp chp kca hai phn t 0 v 1. Khi ta s c tt c dy nh phn c di k.

V d: minh ha dng cy vi k = 3.



#define max 20

int Luu[max];

int k;

/*Xuat ket qua ra man hinh*/

void Out() {

cout


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}

}


void main() {

clrscr();

coutk;

cout


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Bi tp 13. Chnh hp khng lp chp k ca n phn t

Chnh hp chp k ca n phn t ( nk ) l mt nhm c th t gm k phn t khcnhau c chn t n phn t cho.

Phng php: lit k dy c di k v cc phn t trong dy c ly t tp hp {0,

1, , n-1} cc phn t c a vo dy khng c php trng nhau.

V d:n = 3 v k = 2 ta s c cc dy con {0,1}, {0,2}, {1,0}, {1,2}, {2,0} v {2,1}.



#define max 20

char DanhDau[max];

int Luu[max];

int n,k;

/*Khoi tao cac bien*/void Init() {

coutn;

coutk;

//khoi tao tat ca cac dinh chua duoc chon

for(int i = 0; i


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}

}

}


void main() {

clrscr();

cout


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Bi tp 14. Hon v mng s nguyn c n phn t

Phng php: tng t phng php lm bi tp 13 nhng y ta thay tp hp {0, 1, , n-1} l tp hp gi tr n phn t ca mng v di ca dy l n.

V d:n = 3 v A = {-1,0,1} ta s c cc dy con tng ng l {-1,0}, {-1,1}, {0,-1},

{0,1}, {1,-1} v {1,0}.



#define max 20

char DanhDau[max]; //mang danh dau dinh duoc chon

int Luu[max], A[max],n;

/*Khoi tao cac bien*/

void Init() {

coutn;

for(int i = 0; i


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Bi tp 15. t n qun hu trn bn c vua

M t bi ton: lit k tt c phng n t n qun hu trn bn c vua cp nn saocho n qun hu khng c php n nhau.

V d: cho bn c vua cp 88 . Di y l 1 phng n t qun hu:



#define max 20

char a[max]; //danh dau cot

char b[2*max-1]; //danh dau huong Dong-Bac

charc[2*max-1]; //danh dau huong Tay-Bac

int Luu[max]; //luu ket qua tim duoc

int n;

/*Khoi tao cac bien*/

void Init() {

coutn;

//tat ca cac cot chua duoc chon

for(int i = 0; i


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cout


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Bi tp 16. Mi tun

M t bi ton: t qun m ti c v tr (x,y) trn bn c vua cp nn . Hy lit ktt c cc phng n qun m xut pht ti v tr (x,y) c th nhy n tt c cc khc

trn bn c vi iu kin mi qun m ch c php i qua ng 1 ln.

V d: cho bn c vua cp 88 . Ta c 2 phng n t qun m nh sau:

Ci t:#include "iostream.h"

#include "conio.h"

#define max 10

intA[max][max]; //Mang danh dau

int B[max][max]; //Mang luu duong di

intX[8]={-1,-2,-2,-1,1,2,2,1};intY[8]={-2,-1,1,2,2,1,-1,-2};

intn, x, y, Dem=0;

//Khoi tao

void Init() {

coutn;

coutx;

couty;for(int i = 0; i


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void Out() {

Dem++;

for(int i = 0; i


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CHNG 2. SP XP

Bi tp 1. Thut ton Bubble Sort

tng thut ton:xut pht t phn t cui danh sch ta tin hnh so snh vi phn

t bn tri ca n. Nu phn t ang xt c kha nh hn phn t bn tri ca n tatin a n v bn tri ca dy bng cch hon v vi phn t bn tri ca n. Tip tc

thc hin nh th i vi bi ton c n phn t th sau n 1 bc ta thu c danhsch tng dn.

Vd: s dng thut ton Bubble Sort sp xp dy s {3, 10, 4, 6, 2, 6, 15, 3, 9,7} theo

th t tng dn.

Sau 9 bc lp ta thu c dy c sp xp: {2, 3, 3, 4, 6, 6, 7, 9, 10, 15}.

Ci t:#include

#include

#define max 100

//nhap mang

void NhapMang(int A[],int n) {

for(int i=0; i


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//hoan vi 2 phan tu

void Swap(int &a,int &b) {

int temp = a;

a = b;

b = temp;

}

//sap xep cac phan tuvoid BubbleSort(int A[],int n) {

for(int i = 0; i < n-1; i++)

for(int j = n-1; j > i; j--)

if(A[j]


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cout


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Bi tp 2. Thut ton Selection Sort

tng thut ton: xt dy n phn t 110 ,...,, naaa .

Chn trong dy 110 ,...,, naaa ra phn t c kha nh nht v hon v n vi a0.

Chn trong dy 121 ,...,, naaa ra phn t c kha nh nht v hon v n vi a1. C tip tc nh th sau n 1 bc ta thu c danh sch c th t.V d:

Sau 9 bc lp ta thu c dy c sp xp: {2, 3, 4, 5, 6, 6, 7, 7, 8, 9}.

Ci t:#include

#include

#define max 100//nhap mang


for(int i=0; i


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cout


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Bi tp 3. Thut ton Insertion Sort


Xem dy gm 1 phn t a0 l dy c th t. Thm a1 vo dy c th t a0 sao cho dy mi a0, a1 l dy c th t. Nu a1 < a0 ta

hon v a1 vi a0. Thm a2 vo dy c th t a0, a1 sao cho dy mi a0, a1, a2 l dy c th t. Tip tc nh th n n 1 bc ta s c dy c th t 110 ,...,, naaa .V d: s dng thut ton Insertion Sort sp xp dy {3,7,22,3,1,5,8,4,3,9} theo th t

tng dn.

Ci t:#include

#include #define max 100

//nhap mangvoid NhapMang(int A[],int n) {

for(int i=0; i


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Bi tp 4. Thut ton Quick Sort


Bc 1: chn kha 2/RightLeftapivot .

Bc 2: Phn vng. Nhng phn t nh hn kha th nm bn tri ca kha, nhng

phn t ln hn kha th nm bn phi ca kha v nhng phn t bng khac th nm bt c ch no trn dy.

Bc 3: sp xp cho c hai phn vng mi bn tri v bn phi.

M t hot ng ca thut ton Quick Sort:

Ci t:#include #include

#define max 100

void NhapMang(int A[],int n) {for(int i=0; i


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Bi tp 5. Thut ton Heap Sort

Ta xem danh sch n phn t 110 ,...,, naaa l cy nh phn. Cy nh phn ny c xc

nh nh sau: ti nt th i tng ng vi ch s th i ca mng c con tri l nt2*(i+1)-1 v con phi 2*(i+1) nu 2*(i+1)-1 v 2*(i+1) nh hn n. Thut ton c

m t nh sau:- Xy dng Heap sao cho vi mi nt cha u c gi tr ln hn nt con. Khi nt

gc l nt c gi tr ln nht.- Hon v nt gc vi nt th n 1 v xy dng li Heap mi vi n 2 nt v tip

tc hon v nt gc vi nt l cui ca cy mi sau n 2 bc ta s thu c danhsch c sp xp theo th t.

V d: xt danh sch trc khi sp xp0 1 2 3 4 5 6 7

11 3 5 4 9 15 19 7

Danh sch trn c th hin bng cy theo thut ton Heap Sort nh sau:


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Sau 7 bc ta thu c danh sch c sp xp

0 1 2 3 4 5 6 7

3 4 5 7 9 11 15 19

Ci t:#include

#include #define max 100


for(int i=0; i


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Largest = Right;

if(i!=Largest) {

Swap(A[i],A[Largest]);

Heapify(A,n,Largest);

}

}

//xay dung Heap sao cho moi nut cha luon lon hon nut con tren cayvoid BuildHeap(int A[], int n) {

for(int i = n/2-1; i>=0; i--)

Heapify(A,n,i);

}

void HeapSort(int A[], int n) {

BuildHeap(A,n);

for(int i = n-1; i>0; i--){

Swap(A[0],A[i]);

Heapify(A,i,0);

}

}

void main() {

int A[max], n;

clrscr();

coutn;

NhapMang(A,n);

cout


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Bi tp 6. Thut ton Merge Sort

M t bi ton: cho 2 danh sch A v B ln lt c m v n phn t sp xp theo th

t. Bi ton t ra trn 2 danh sch A v B vi nhau thnh danh sch C cng l mtdanh sch c th t.

Thut ton:

Bc 1: khi to ba ch s chy trong vng lp i = 0, j = 0, k = 0 tng ng cho bamng A, B v C.

Bc 2: ti mi bc nu c hai ch s (i


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k++;

}

if (i < m) {

for (int p = i; p < m; p++) {

C[k] = A[p];

k++;

}} else {

for (int p = j; p < n; p++) {

C[k] = B[p];

k++;

}

}

}

void main() {

int A[max],B[max],C[max],n,m,k;

clrscr();

coutn;

coutm;

cout


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CHNG 3. I S MA TRN

Bi tp 1. Nhp xut ma trn

Phng php: ta t chc mng hai chiu lu tr cc phn t trong ma trn.Ci t:#include "conio.h"


#define max 100

/*Nhap ma tran he so*/

void NhapMaTran(float A[max][max], int m, int n) {

for(int i = 0; i


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Bi tp 2. Mt s php ton trn ma trn

Cng ma hai ma trnCho ijaA v ijbB l hai ma trn cng cp nm . Khi BAC cng l

ma trn cp nm v c xc nh bi njmibac ijijij 1,1, .

Nhn hai ma trnCho ijaA l ma trn cp nm v jkbB l ma trn cp pn . Khi

ikcBAC . l ma trn cp pm v c xc nh nh bi

n

jjkijik bac

1

vi

mi ...,,1 v pk ...,,1 .

Ci t:#include "conio.h"#include "iostream.h"

#define max 100/*Nhap ma tran he so*/

void NhapMaTran(float A[max][max], int m, int n, char ch) {

for(int i = 0; i


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Bi tp 3. H phng trnh tuyn tnh dng tam gic trn

Cho h phng trnh tuyn tnh dng tam gic trn:

B

n

X

n

A

nn

n

n

b

b

b

x

x

x

a

aa

aaa

......

.....00

........

.....0

.....

2

1

2

1

222

11211

niaaaaA iinn ,...,2,1,00...det 2211 .

Nghim ca h trn c xc nh nh sau:

1,...,2,1,1

,...,2,1,

1

nnixaba

x

a

bx

nibxa

BAX

n

ijjiji

iii

nn

nn

i

n

ijjij



#define max 100//Nhap ma tran tam giac tren

void Nhap(float A[max][max],int n) {

for(int i = 0; i


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cout


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coutn;

cout


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for(int j = 0; j


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else {

X[i] = B[i];

for(int j=0; j


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Bi tp 5. Thut ton phn r ma trn A = LU

Qu trnh chuyn ho ma trn A ban u thnh tch hai ma trn tam gic UL. da vophp kh Gauss c thc hin bng cc php nhn ma trn. Thut ton ny c gil thut ton Crout. Qu trnh Crout bao gm nhiu bc hi quy. Nu ma trn A c

cp nn ta cn 1n bc. Thut ton c th hin c th nh sau:

nnT

T

n

T

n

a

a

aCrout

aULvwH

w

IvHv

w

A

1

11

11

1111

110

01

1

111

11

2

222

11211

1

21

0

01

.....0

...........

.....0

.....

1.....0

...........

0.....1

0.....01

n

T

n

nnn

n

n

n

A

uul

aa

aa

uuu

l

l

I

1

111

1111

111

112 0

01

0

01

n

T

nnn

T

n

uu

l

uu

lCrout

ULULI

....

00......

00

....0

....

......

0.....01

0.....001

2

222

11211

21

2

3231

21

Crout

uu

uuu

ll

ll

l

n

n

n

nn

n

AI

UL.

....00

..........

....0

...

1.....

...........

0.....1

0.....01

222

11211

21

21

1

nn

n

n

nn

n

u

uu

uuu

ll

lCrout

Thng qua 1n bc hi quy ta c ULA . vi L l ma trn tam gic di v Ul ma trn tam gic trn.Ci t:

#include "conio.h"#include "iostream.h"

#define max 100


for(int i = 0; i


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}

}

void XuatMaTran(float A[max][max], int n) {

cout


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Bi tp 6. Gii h phng trnh tuyn tnh da vo phn r LU

tng thut ton: cho h phng trnh tuyn tnh tng qut BXA . . Ta tin hnhphn r ULA . . Trong , L l ma trn tam gic di v U l ma trn tam gic trn.

Khi ,

YXUBYLBXULBXA

.....



#define max 100

/*Nhap ma tran he so*/


for(int i = 0; i


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L[i][k] = A[i][k]/U[k][k];

U[k][i] = A[k][i];

U[i][k] = 0;

L[k][i] = 0;

}

for(i = k+1; i


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Bi tp 7. nh thc ca ma trn

tng thut ton: ta tin hnh phn r ma trn ULA . .

Ta c: Det(A)=Det(L)*Det(U) m Det(L) = 1 nn Det(A) = Det(U).

Ci t:

#include "conio.h"#include "iostream.h"

#define max 100

/*Nhap ma tran*/


for(int i = 0; i


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CHNG 4. MT S THUT GII TRN TH

Bi tp 1 . Xt tnh lin thng ca th

M t bi ton: cho th v hng G=(V,E) hy kim tra tnh lin thng ca th G.

tng thut ton:

Bc 1:xut pht t mt nh bt k ca th. Ta nh du nh xut pht v chuynsangBc 2.

Bc 2: t mt nh i nh du, ta nh du nh j nu A[i,j] = 1 v j cha c nhdu v chuyn sangBc 3.

Bc 3: thc hin Bc 2 cho n khi khng cn thc hin c na chuyn sangBc 4.

Bc 4: kim tra nu snhnh du nh hn n (hay tn ti t nht mtnh cha cnh du) th s khng lin thng v ngc li th lin thng.

M t d liu u vo v u ra ca bi ton:D liu vo: cho trong tp tin Bai1.inp

- Dng u ghi s n l s nh ca mt th (0


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Ci t:#include

#include

#include

#define TenFile "Bai1.inp"

//doc du lieu tu tap tin

void Doc_File(int **A,int &n) {

FILE*f = fopen(TenFile,"rb");

fscanf(f,"%d",&n);

*A = new int [n];

cout


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}

}

}while (ThanhCong == 0); //lap lai cho den khi khong con kha nang loan

return 0;

}

//chuong trinh chinh

void main(){

clrscr();

int **A,n;

Doc_File(A,n);

if (Lien_Thong(A,n)==1)

cout


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Bai2.inp OUTPUT

6

0 1 1 0 0 0

1 0 1 0 0 0

1 1 0 0 0 0

0 0 0 0 1 1

0 0 0 1 0 1

0 0 0 1 1 0

THANH PHAN

LIEN THONG: 2

Bai2.inp OUTPUT

8

0 1 0 0 0 1 0 0

1 0 1 0 0 0 0 0

0 1 0 1 0 0 0 0

0 0 1 0 1 0 0 0

0 0 0 1 0 1 0 0

1 0 0 0 1 0 1 0

0 0 0 0 0 1 0 1

0 0 0 0 0 0 1 0

THANH PHAN

LIEN THONG: 1

Ci t:#include

#include

#include

#define TenFile "Bai2.inp"



FILE*f = fopen(TenFile,"rb");

fscanf(f,"%d",&n);

*A = new int [n];

cout


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int Dem=0, i,j, MLT=0;

for( i = 0; i


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Bi tp 3. Tm mi ng i t gia hai nh

M t bi ton: cho th v hng G=(V,E) hy xc nh mi ng i t nh D tinh C ca th G.

tng thut ton: s dng k thut tm kim theo chiu su.

M t d liu u vo v u ra ca bi ton:D liu vo: th lin thng v cho trong tp tin Bai3.inp



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FILE*f = fopen(FileIn,"rb");

fscanf(f,"%d%d%d",&n,&D,&C);

cout


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L[SoCanh] = 0;

DanhDau[i] = 0; //phuc hoi dinh da di qua

}

}

}

//chuong trinh chinh

void main() {Doc_File();

KhoiTao();

cout


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Bai4.inp OUTPUT

5

1

0 1 0 0 0

1 0 1 1 1

0 1 0 1 0

0 1 1 0 1

0 1 0 1 0

12345

Ci t:#include

#include

#include

#define FileIn "Bai4.inp"

int Dem = 0; //dem so duong di

int *L; //luu duong di

char *DanhDau; //danh dau dinh da diint **A,n,D;

//doc du lieu tu tap tin theo yeu cau cua bai toan

void Doc_File() {

FILE*f = fopen(FileIn,"rb");

fscanf(f,"%d%d",&n,&D);

cout


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L[0] = D;

}

//xuat duong di ra man hinh

void InDuongDi(int Dinh) {

Dem++;

cout


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Bi tp 5. ng i Euler

M t bi ton: cho th v hng G=(V,E) hy xc nh mi ng i qua tt c cccnh mi cnh ch qua duy nht 1 ln.

tng thut ton: s dng k thut tm kim theo chiu su bng cch xa cnh

i qua trong qu trnh tm kim ng i.

M t d liu u vo v u ra ca bi ton:D liu vo: cho trong tp tin Bai5.inp

- Dng u ghi s n l s nh ca mt th (03->4->5->2->1

Ci t:#include

#include

#include #define Filename "Bai5.inp"

int Dem = 0, SoCanh=0; //dem so duong di va luu so canh cua do thi

int *L; //luu dinh da di quaint **A,n;

int XuatPhat=0; //dinh xuat phat la dinh bac le neu do thi co dinh bac le

void Doc_File() {

int BacDinh;

FILE*f = fopen(Filename,"rb");

fscanf(f,"%d",&n);

cout


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if(A[i][j] == 1)

BacDinh++;

}

if(BacDinh%2==1)

XuatPhat = i; //xuat phat tu dinh bac le

SoCanh+=BacDinh;

cout


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Bi tp 6. Thut ton Dijkstra tm ng i ngn nht

M t bi ton: cho th v hng G=(V,E) hy xc nh ng i ngn nht t nhD ti nh C ca th G.

tng thut ton: s dng thut ton Dijkstra.

M t d liu u vo v u ra ca bi ton:D liu vo: th lin thng v cho trong tp tin Bai6.inp.



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Bi tp 7. Thut ton Prim tm cy bao trm ti tiu

M t bi ton: cho th v hng c trng s G=(V,E) hy tm ng i sao cho ttc cc nh iu c ng i vi nhau v tng trng s ca ng i l nh nht. Tc l

tm th con lin thng GG ' sao cho tng trng s ca G l nh nht.

tng thut ton:

Bc 1: xut pht t nh k bt k (thng thng chn nh u tin) chn mt cnh ctrng s nh nht lin k vinh k (min{A[k][j]}j=1..n) ta nh du 2 nh i quacnh v s cnh tm c l 1. Chuyn sangBc 2.

Bc 2: tm cnh nh nht ca th viiu kin cnh tm c phi c 1 nh cha

nh du v 1 nh nh du. Ngha l, ta tm min{A[i][j]}j=1..n, i=1..n saocho i nhu v j cha nh du trnh trng hp to thnh chu trnh

con. Ta tng s cnh tm c ln 1 v chuyn sangBc 3.

Bc 3: nu s cnh tm c bng n-1 kt thc thut ton, ngc li quay vBc 2.

M t d liu u vo v u ra ca bi ton:D liu vo: lu trong tp tin Bai7.inp



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void Doc_File(int A[max][max],int &n) {

FILE*f = fopen(FileInt,rb);

fscanf(f,%d,&n);

for(int i =0;i


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L[Dem].x = i;

L[Dem].y = j;

}

Sum+=min;

D[L[Dem].y] = 1;

Dem++;

} while(Dem


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V d:

Bai8.inp Bai8.out8

0 2 3 4 0 0 0 0

2 0 3 0 4 0 0 0

3 3 0 7 6 5 2 0

4 0 7 0 0 0 3 00 4 6 0 0 4 0 8

0 0 5 0 4 0 1 6

0 0 2 3 0 1 0 5

0 0 0 0 8 6 5 0

20

6 - 7

1 - 2

3 - 7

1 - 34 - 7

2 - 5

7 - 8

Ci t:#include

#include

#define FileInt "Bai8.inp"#define FileOut "Bai8.out"

typedef struct Egde {

int x,y;

};



FILE*f = fopen(FileInt,"rb");

fscanf(f,"%d",&n);

*A = new int [n];

for(int i =0;i


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int min = MAXINT, Dem = 0, Sum = 0, T = 0, Temp;

for(int i=0; i


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TI LIU THAM KHO

[1] Chu c Khnh, Nguyn Cam, L thuyt th, NXB i hc Quc giaTPHCM.

[2]

Dng Thy V, Gio trnh Phng php tnh, NXB KHKT.[3] Ngc Anh Th, Gio trnh thut ton, NXB Thng k.[4] Ng c Tn,L thuyt t hp v th, NXB H Quc gia H Ni.[5] Dng V Diu Tr, Nguyn Tin Huy, Trn an Th, V Mnh Trng, Cm

nang thut ton, NXB KHKT.

[6] Xun Li, Cu trc d liu & Gii thut, 1995.[7] Nguyn c Ngha, T Vn Thnh, Ton ri rc, 1997.[8] A.V. Aho, J.E. Hopcroft, J.D. Ullman, Data Structures and Algorithms, Addison

Wesley, 1983.

[9] Jeffrey H Kingston,Algorithms and Data Structures, Addison-Wesley, 1998.

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