华为面试题：迷宫问题 C语言源码

定义一个二维数组N*M（其中2<=N<=10;2<=M<=10），如5 × 5数组下所示：

int maze[5][5] = {

0, 1, 0, 0, 0,

0, 1, 0, 1, 0,

0, 0, 0, 0, 0,

0, 1, 1, 1, 0,

0, 0, 0, 1, 0,

};

它表示一个迷宫，其中的1表示墙壁，0表示可以走的路，只能横着走或竖着走，不能斜着走，要求编程序找出从左上角到右下角的最短路线。入口点为[0,0],既第一空格是可以走的路。

Input

一个N × M的二维数组，表示一个迷宫。数据保证有唯一解,不考虑有多解的情况，即迷宫只有一条通道。

Output

左上角到右下角的最短路径，格式如样例所示。

5 5

0 1 0 0 0

0 1 0 1 0

0 0 0 0 0

0 1 1 1 0

0 0 0 1 0

输出

(0,0)

(1,0)

(2,0)

(2,1)

(2,2)

(2,3)

(2,4)

(3,4)

(4,4)

#include "stdio.h"
#include "stdlib.h"
#include "string.h"
#define MAX_PATH 256

int maze[10][10] = {0};
int route[100][2] = {0};
int main()
{
	int row=0,line=0;
	scanf("%d %d",&row,&line);
	for (int i=0;i<row;i++)
	{
		for (int j=0;j<line;j++)
		{
			scanf("%d",&maze[i][j]);
		}
	}
	//走迷宫
	//堆栈:记录上一个位置
	int xcurrent = 0;
	int ycurrent = 0;
	int count=0;
	while(true)
	{
		if (maze[xcurrent+1][ycurrent]==0 && xcurrent+1<row)
		{
			//返回上一个位置
			if (route[count-1][0]==xcurrent+1 && route[count-1][1]==ycurrent)
			{
				maze[xcurrent][ycurrent]=1;//设置为墙
				count--;
				xcurrent++;
			}
			else
			{
				route[count][0]=xcurrent;
				route[count][1]=ycurrent;
				count++;
				xcurrent++;
			}
		}
		else if (maze[xcurrent][ycurrent+1]==0 && ycurrent<line)
		{
			if (route[count-1][0]==xcurrent && route[count-1][1]==ycurrent+1)
			{
				maze[xcurrent][ycurrent]=1;//设置为墙
				count--;
				ycurrent++;
			}
			else
			{
				route[count][0]=xcurrent;
				route[count][1]=ycurrent;
				count++;
				ycurrent++;
			}
		}
		else if (maze[xcurrent-1][ycurrent]==0 && xcurrent-1>=0)
		{
			if (route[count-1][0]==xcurrent-1 && route[count-1][1]==ycurrent)
			{
				maze[xcurrent][ycurrent]=1;//设置为墙
				count--;
				xcurrent--;
			}
			else
			{
				route[count][0]=xcurrent;
				route[count][1]=ycurrent;
				count++;
				xcurrent--;
			}
		}
		else if (maze[xcurrent][ycurrent-1]==0 && ycurrent-1>=0)
		{
			if (route[count-1][0]==xcurrent && route[count-1][1]==ycurrent-1)
			{
				maze[xcurrent][ycurrent]=1;//设置为墙
				count--;
				ycurrent--;
			}
			else
			{
				route[count][0]=xcurrent;
				route[count][1]=ycurrent;
				count++;
				ycurrent--;
			}
		}
		if (xcurrent==row-1 && ycurrent==line-1)
		{
			route[count][0]=xcurrent;
			route[count][1]=ycurrent;
			count++;
			break;
		}
	}

	for (int i=0;i<count;i++)
	{
		printf("(%d,%d)\n",route[i][0],route[i][1]);
	}

	return 0;
}