3.6 用栈实现解决迷宫问题

程序代码:

#include<stdio.h>
#define M 4
#define N 4
#define MaxSize 100

int mg[M+2][N+2]=
{
	{1,1,1,1,1,1},
	{1,0,0,0,1,1},
	{1,0,1,0,0,1},
	{1,0,0,0,1,1},
	{1,1,0,0,0,1},
	{1,1,1,1,1,1}
};

struct 
{
	int i,j;
	int di;
}St[MaxSize],Path[MaxSize];


int top = -1;
int count =1;
int minlen = MaxSize;

void dispapath()
{
	int k;
	printf(" %5d: ",count++);
	for(k=0;k<=top;k++)
		printf("(%d,%d)",St[k].i,St[k].j);
	printf("\n");
	if(top+1 <minlen)
	{
		for(k=0;k<=top;k++)
			Path[k]=St[k];
		minlen = top+1;
	}
}

void dispminpath()
{
	printf("最短路径如下:\n");
	printf("长度:%d\n",minlen);
	for(int k=0;k<minlen;k++)
		printf("(%d,%d)",Path[k].i,Path[k].j);
	printf("\n");
	
}

void mgpath(int xi,int yi,int xe,int ye)
{
	int i,j,i1,j1,di;
	bool find;
	top++;
	St[top].i =xi;
	St[top].j =yi;
	St[top].di=-1;
	mg[xi][yi]=-1;


while(top>-1)
{
	i= St[top].i;
	j= St[top].j;
	di = St[top].di;
	if(i==xe&&j==ye)
	{
		dispapath();
		mg[i][j]=0;
		top--;
		i = St[top].i;
		j = St[top].j;
		di = St[top].di;
	}
	
	find =false;
	while(di<4&&!find)
	{
		di++;
		switch(di)
		{
			case 0:i1 =i-1;j1=j;break;
			case 1:i1=i;j1=j+1;break;
			case 2:i1= i+1;j1= j;break;
			case 3:i1=i,j1 =j-1;break;
		}
		
		if(mg[i1][j1]==0) find=true;
	}
	
	if(find)
	{
		St[top].di =di;
		top++;
		St[top].i= i1;
		St[top].j= j1;
		St[top].di= -1;
		mg[i1][j1]=-1;
	}
	else
	{
		mg[i][j]=0;
		top--;
	}
}
dispminpath();

}

int main()
{
	printf("迷宫所有路径如下:\n");
	mgpath(1,1,M,N);
	return 1;
}

运行截图:

《3.6 用栈实现解决迷宫问题》

    原文作者:迷宫问题
    原文地址: https://blog.csdn.net/wuwuhuai/article/details/79281843
    本文转自网络文章,转载此文章仅为分享知识,如有侵权,请联系博主进行删除。
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