迷宫问题——回溯法解

题目描述

    迷宫是一个二维矩阵,其中1为墙,0为路,入口在第一列,出口在最后一列。
    要求从入口开始,从出口结束,按照 上,下,左,右 的顺序来搜索路径.

输入

    第一行输入迷宫大小N
    第二行输入入口坐标
    接下来n行输入完整迷宫

输出

    输出完整棋盘,所有解法,走过的路用6表示。

样例输入

 8 
 0 7
 1 1 1 1 1 1 1 1
 1 0 1 1 0 0 0 0
 1 0 1 0 0 1 0 1
 1 1 0 0 1 0 1 1
 1 0 0 1 0 0 0 1
 1 0 0 0 0 1 1 1
 1 0 1 0 0 1 0 1
 0 0 1 0 0 0 1 1
 1 1 1 1 0 0 0 1
 1 1 1 1 1 1 1 1 

样例输出

 1 1 1 1 1 1 1 1
 1 0 1 1 6 6 6 6
 1 0 1 6 6 1 0 1
 1 1 6 6 1 0 1 1
 1 6 6 1 0 0 0 1
 1 6 0 0 0 1 1 1
 1 6 1 0 0 1 0 1
 6 6 1 0 0 0 1 1
 1 1 1 1 0 0 0 1
 1 1 1 1 1 1 1 1

实现代码

#include <stdio.h>
#define N 19 //整个迷宫大小(包括最外层墙壁)

char Maze[N][N] = {
    {1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1},
    {1,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0},
    {1,1,1,0,1,1,1,0,1,0,1,1,1,1,1,0,1,1,1},
    {1,0,0,0,0,0,1,0,1,0,0,0,0,0,1,0,0,0,1},
    {1,1,1,0,1,0,1,0,1,1,1,1,1,0,1,1,1,1,1},
    {1,0,1,0,1,0,1,0,1,0,0,0,1,0,0,0,0,0,1},
    {1,0,1,0,1,1,1,0,1,1,1,0,1,1,1,1,1,0,1},
    {1,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,1},
    {1,0,1,1,1,1,1,1,1,1,1,0,1,0,1,1,1,1,1},
    {1,0,1,0,0,0,0,0,0,0,1,0,1,0,1,0,0,0,1},
    {1,0,1,0,1,1,1,0,0,0,1,0,1,0,1,0,1,0,1},
    {1,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,1,0,1},
    {1,0,1,1,1,0,1,1,1,1,1,1,1,0,1,1,1,0,1},
    {1,0,1,0,1,0,0,0,0,0,0,0,1,0,1,0,0,0,1},
    {1,0,1,0,1,1,1,1,1,1,1,0,1,0,1,0,1,0,1},
    {1,0,1,0,1,0,0,0,1,0,0,0,1,0,1,0,1,0,1},
    {1,0,1,0,1,0,1,0,1,0,1,1,1,1,1,0,1,0,1},
    {0,0,1,0,0,0,1,0,0,0,0,0,0,0,0,0,1,0,1},
    {1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1}
};
char step[4][2] = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};   //定义四个方向
int count = 0;

char Check(char i, char j)  //判断下一步是不是通路
{
    if(i >= 0 && i<=18 && j >= 0 && j <= 18)
    {
        if(0 == Maze[i][j])
        {
            return 1;
        }
    }
    return 0;
}

void Display(void)
{
    char i, j;

    for(i=0; i<N; ++i)
    {
        for(j=0; j<N; ++j)
        {
            printf("%d", Maze[i][j]);
        }
        printf("\n");
    }
}

void Find(char ci, char cj)
{
    char n;
    if((N-1) == cj) //这个边界设置得比较粗糙,到N-1这个下标就算终点,没想到什么好的办法
    {
        Maze[ci][cj] = 6;   //终点的最后一个6
        printf("解法%d(Enter回车查看下一解法):\n", ++count);
        Display();
        getchar();
        Maze[ci][cj] = 0;
    }
    else
    {
        for(n=0; n<4; ++n)
        {
            if(Check(ci+step[n][0], cj+step[n][1])) //依然用Check实现判断:是否可以在某个方向走下一步
            {
                Maze[ci][cj] = 6;   //6表示走过的路
                Find(ci+step[n][0], cj+step[n][1]);
                Maze[ci][cj] = 0;   //传统回溯方法
            }
        }
    }
}

int main(void)
{
    printf("\t迷宫问题(1 墙壁 0 道路 6 解法)\n\n");
    printf("原迷宫(Enter回车查看解法):\n");
    Display();
    getchar();

    //(17, 0)为起点
    Find(17, 0);

    return 0;
}
    原文作者:回溯法
    原文地址: https://blog.csdn.net/zhuofeilong/article/details/48129995
    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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