迷宫问题——回溯法解

来源：互联网发布：top域名能备案吗编辑：程序博客网时间：2024/05/21 07:04

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;}

7 0