算法: Mazing-老鼠走迷宫

说明

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老鼠走迷宫是递回求解的基本题型.咱们在二维阵列中使用 2 表示迷宫墙壁,使用 1 来表示老鼠的行走路径,尝试以程式求由入口至出口的路径.

解法

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老鼠可走上,下,左,右四个方向.入口通常在左上,出口在右下(若是说有其余出入口状况则下述老鼠每步默认选择方向的顺序改变)

老鼠从入口出发,每走一步进行一次四个方向的选择,判断下步是否有可走方向(因为入口在左上,出口在左下,则优先选择右,下方向,以后选择左,上方向).没法前进时退回上步选择下一个可前进方向.

如此在阵列中按照顺序测试四个方向,直到走出出口为止.

代码(c语言版,思路可参考)

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参数声明

int maze[7][7] = { {2,2,2,2,2,2,2},
		   {2,0,0,0,0,0,2}, 
	           {2,0,2,0,2,0,2}, 
	           {2,0,0,2,0,2,2}, 
	           {2,2,0,2,0,2,0}, 
	           {2,0,0,0,0,0,2}, 
	           {2,2,2,2,2,2,2} };//mazing map
	int(*mazeP)[7];
	mazeP = maze;//将二维数组传递给数组指针(数组指针是指针指向一个数组)

	int startI = 1, startJ = 1, endI = 5, endJ = 5;//entrance and exit
        int success = 0;//success mark
	int i = 1,j = 1;//initialization of import and export
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递回方法

//递回进行老鼠每步选择的方向,返回是否成功
int mazing_visit(int i, int j,int(*mazeP)[7],int success)
{
	mazeP[i][j] = 1;

	if (i == 5 && j == 5)success = 1;

	//优先右下,而后左上
	if ( 1 != success && mazeP[i][j + 1] == 0)mazing_visit(i, j + 1, mazeP, success);
	if (success != 1 && mazeP[i + 1][j] == 0)mazing_visit(i + 1, j, mazeP, success);
	if (success != 1 && mazeP[i][j - 1] == 0)mazing_visit(i, j - 1, mazeP, success);
	if (success != 1 && mazeP[i - 1][j] == 0)mazing_visit(i - 1, j, mazeP, success);
		

	if (success != 1)mazeP[i][j] == 0;

	return success;
}
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完整代码

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
//Mazing-老鼠走迷宫
int mazing_visit(int, int,int (*p)[7],int);

void mazing()
{
	int maze[7][7] = { {2,2,2,2,2,2,2},
						{2,0,0,0,0,0,2}, 
	                    {2,0,2,0,2,0,2}, 
	                    {2,0,0,2,0,2,2}, 
	                    {2,2,0,2,0,2,0}, 
	                    {2,0,0,0,0,0,2}, 
	                    {2,2,2,2,2,2,2} };//mazing map
	int(*mazeP)[7];
	mazeP = maze;//将二维数组传递给数组指针(数组指针是指针指向一个数组)

	int startI = 1, startJ = 1, endI = 5, endJ = 5;//entrance and exit
        int success = 0;//success mark
	int i = 1,j = 1;//initialization of import and export

	printf("显示迷宫: \n");
	for (i = 0; i < 7; i++)
	{
		for (j = 0; j < 7; j++)
		{
			if (mazeP[i][j] == 2)//判断是不是墙壁
			{
				printf("*");
			}
			else
			{
				printf(" ");
			}
		}
		printf("\n");
	}
	
        //判断是否存在出口,如有出口则打印出路径
	if (mazing_visit(startI, startJ, mazeP,success) == 0)
	{
		printf("\n没有找到出口\n");
	}
	else
	{
		printf("\n显示路径:");
		for (i = 0; i < 7; i++)
		{
			for (j = 0; j < 7; j++)
			{
				if (mazeP[i][j] == 2)
				{
					printf("*");
				}
				else if (mazeP[i][j] == 1)
				{
					printf(">");
				}
				else
				{
					printf(" ");
				}
			}
			printf("\n");
		}
			
	}

}

int mazing_visit(int i, int j,int(*mazeP)[7],int success)
{
	mazeP[i][j] = 1;

	if (i == 5 && j == 5)success = 1;

	//优先右下,而后左上
	if ( 1 != success && mazeP[i][j + 1] == 0)mazing_visit(i, j + 1, mazeP, success);
	if (success != 1 && mazeP[i + 1][j] == 0)mazing_visit(i + 1, j, mazeP, success);
	if (success != 1 && mazeP[i][j - 1] == 0)mazing_visit(i, j - 1, mazeP, success);
	if (success != 1 && mazeP[i - 1][j] == 0)mazing_visit(i - 1, j, mazeP, success);
		

	if (success != 1)mazeP[i][j] == 0;

	return success;
}
int main(void)
{
    mazing();
    return 0;
}
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