迷宫求解（C语言回溯法）

#include "stdio.h"

struct stack

{

int h;

struct node *top;

};typedef struct stack STK;

struct node

{

int x;

int y;

struct node *next;

};typedef struct node STN;

void dayin(STK *p)

{

STN *q;

q=(STN*)malloc(sizeof(STN));

q->next=p->top;

q=q->next;

while(q->next!=NULL)

{

printf("(%d,%d)",q->x,q->y);

q=q->next;

}

printf("(%d,%d)",q->x,q->y);

}

pop(STK *a,int *b,int *c)

{

STN *o;

o=(STN *)malloc(sizeof(STN));

o->next=a->top;

o=o->next;

*b=o->x;

*c=o->y;

a->top=o->next;

a->h--;

free(o);

}

dayinmigong(int (*u)[5])

{

int a,b;

for(a=0;a<6;a++)

{

for(b=0;b<5;b++)

{

if(u[a][b]==1)printf(" ");

else if(u[a][b]==0)printf("#");

else if(u[a][b])printf("o");

}

printf("\n");

}

}

tot(STK *a,int *b,int *c)

{

STN *o;

o=(STN *)malloc(sizeof(STN));

o->next=a->top;

o=o->next;

*b=o->x;

*c=o->y;

}

tot2(STK *a,int *b,int *c)

{

STN *o;

o=(STN *)malloc(sizeof(STN));

o->next=a->top;

o=o->next;

o=o->next;

*b=o->x;

*c=o->y;

}

main()

{

STK *a;

STN *p;

int a3=0,b3=0;

int a1=99,b1=99,a2=99,b2=99;

int b[6][5]={1,0,1,1,1,1,0,1,1,1,1,1,1,0,1,1,1,0,0,1,1,1,0,1,1,1,1,0,1,2};

a=(STK*)malloc(sizeof(STK));

a->h=0;

p=(STN*)malloc(sizeof(STN));

a->top=p;

p->next=NULL;

p->x=0;

p->y=0;

a->h++;

dayinmigong(b);

while(b[a3][b3]!=2)

{

if(((b3+1)<5)&&(b[a3][b3+1]==1||b[a3][b3+1]==2)&&((b3+1)!=b2||(a3!=b2))&&((b3+1)!=b1||a3!=a1))

{

b3=b3+1;

p=(STN*)malloc(sizeof(STN));

p->next=a->top;

a->top=p;

p->x=a3;

p->y=b3;

a->h++;

tot2(a,&a2,&b2);

}

else if(((a3+1)<6)&&(b[a3+1][b3]==1||b[a3+1][b3]==2)&&((a3+1)!=a2||(b3!=b2))&&((a3+1)!=a1||(b3!=b1)))

{

a3=a3+1;

p=(STN*)malloc(sizeof(STN));

p->next=a->top;

a->top=p;

p->x=a3;

p->y=b3;

a->h++;

tot2(a,&a2,&b2);

}

else if(((b3-1)>0)&&(b[a3][b3-1]==1||b[a3][b3-1]==2)&&((b3-1)!=b2||(a3!=a2))&&((b3-1)!=b1||(a3!=a1)))

{

b3=b3-1;

p=(STN*)malloc(sizeof(STN));

p->next=a->top;

a->top=p;

p->x=a3;

p->y=b3;

a->h++;

tot2(a,&a2,&b2);

}

else if(((a3-1)>0)&&(b[a3-1][b3]==1||b[a3-1][b3]==2)&&((a3-1)!=a2||(b3!=b2))&&((a3-1)!=a1||(b3!=b1)))

{

a3=a3-1;

p=(STN*)malloc(sizeof(STN));

p->next=a->top;

a->top=p;

p->x=a3;

p->y=b3;

a->h++;

tot2(a,&a2,&b2);

}

else if(b[a3][b3]!=2)

{

pop(a,&a1,&b1);

tot(a,&a3,&b3);

if(a3==0&&b3==0)continue;

tot2(a,&a2,&b2);

}

}

dayin(a);

scanf("%d");

}