数据结构4——迷宫问题（栈）

利用栈的思想，解决迷宫问题：
我们通过一个二维数组定义一个迷宫

int migong[7][7]={
{1,1,1,1,1,1,1}, 
{1,0,0,1,0,0,1},
{1,0,0,1,0,1,1},
{1,0,0,0,0,0,1},
{1,0,1,0,1,0,1},
{1,0,1,0,1,0,1},
{1,1,1,1,1,1,1}
};

以下解决迷宫问题

#include<stdio.h>
#include<stdlib.h>
typedef int elmtype;
typedef struct{
	elmtype row;
	elmtype column;
	elmtype derection;
}Lu;
typedef struct{
	Lu date[50];
	int top;
}SequenStack;
SequenStack *init();
int Push(SequenStack *S,Lu t);
int Pop(SequenStack *S,Lu *t);
int Get(SequenStack *S,Lu *t);
int main(){
	int migong[7][7]={
{1,1,1,1,1,1,1}, 
{1,0,0,1,0,0,1},
{1,0,0,1,0,1,1},
{1,0,0,0,0,0,1},
{1,0,1,0,1,0,1},
{1,0,1,0,1,0,1},
{1,1,1,1,1,1,1}
};
	SequenStack *s;
	Lu temp;
	Lu tryPath;
	Lu t;
	int x,y,d,i,j,time,yet=0;
	s=init();
	temp.row=1;
	temp.column=1;
	temp.derection=0;
	Push(s,temp);
	while(s->top!=-1)
	{
		Get(s,&temp);
		x=temp.row;
		y=temp.column;
		d=temp.derection+1;
		time=0;
		while(d<=4)
		{
			if(d==1){i=x;j=y+1;}
			if(d==2){i=x+1;j=y;}
			if(d==3){i=x;j=y-1;}
			if(d==4){i=x-1;j=y;}
			if(migong[i][j]==0)
			{
				if(time==0)
				{
					Pop(s,&tryPath);
					tryPath.derection=d;
					Push(s,tryPath);
					time++;
				}
			temp.row=i;
			temp.column=j;
			temp.derection=1;
			Push(s,temp);
			x=i;
		    y=j;
		    migong[x][y]=-1;
		    if(x==5&&y==5)
			    yet=1;
			    else
			    	d=1;
			}
			else
			{
				d++;
				time=0;
			}
			if(d==5&&s->top!=-1)
			{
				Pop(s,&temp);
				migong[x][y]='X';
			}
			if(s->top==-1)
			yet=0;
		}
		int i=0;
	if(yet==1){
		printf("该迷宫的路径为\n");
		while(i<s->top){
			printf("%d,%d->",s->date[i].row,s->date[i].column);
			i++;
		}
		if(i==s->top)
		{
			printf("%d,%d",s->date[i].row,s->date[i].column);
		}
	}return 1;
	}
}	
SequenStack *init(){
	SequenStack *S;
	S=(SequenStack*)malloc(sizeof(SequenStack));
	S->top=-1;
	return S; 
}
int Push(SequenStack *S,Lu t){
	S->top++;
	S->date[S->top]=t;
	return 1;
}
int Pop(SequenStack *S,Lu *t){
	S->top--;
	*t=S->date[S->top+1];
	return 1;
}
int Get(SequenStack *S,Lu *t){
	*t=S->date[S->top];
}