广度遍历实现迷宫求解

  运用队列迷宫求解。

#include <stdio.h>
#include "stdlib.h"
#define MAXSIZE 150

int mg2[10][10] = {
                {1,1,1,1,1,1,1,1,1,1},
                {1,0,0,1,0,0,0,1,0,1},
                {1,0,0,1,0,0,0,1,0,1},
                {1,0,0,0,0,1,1,0,0,1},
                {1,0,1,1,1,0,0,0,0,1},
                {1,0,0,0,1,0,0,0,0,1},
                {1,0,1,0,0,0,1,0,0,1},
                {1,0,1,1,1,0,1,1,0,1},
                {1,1,0,0,0,0,0,0,0,1},
                {1,1,1,1,1,1,1,1,1,1}
};

struct node
{
    int i,j,pre,di;
}sq[MAXSIZE];

void print(int front)
{   int pre = sq[front].pre;
    printf("倒着输出的路径为: \n");
    while(pre != -1)
    {
        printf("i = %d ,j = %d ,di = %d\n",sq[pre].i, sq[pre].j, sq[pre].di);
        pre = sq[pre].pre;
    }
    printf("\n\n");
}

void mgpath2(int x1, int y1, int x2, int y2)
{

    int front = -1; int rear = -1; int find = 0;  int i,j;     //front rear 是队列的头部和尾部
    rear++;
    sq[rear].i = i = x1; sq[rear].j = j = y1; sq[rear].di = -1; sq[rear].pre = -1;  //把开始的节点进队列，并初始化前一节点的下标和方向为 -1
    mg2[i][j] = -1;
    while(front <= rear)     //当队列中还有元素的时候进行循环
    {
        front ++ ;
        find = 0;              //出队列
        if(sq[front].i == x2 && sq[front].j == y2)   //判断到达目标的时候，打印输出路径
        {
            printf(front);              //此处没有直接返回 ，是想要打印全部的路径，但是目的没有达到，因为走过的节点被赋值为-1，也就不能再走一次，所以不能实现打印所有路径
        }
        while(sq[front].di < 4 && find == 0)      //此处得循环必须是四次，因为要把周围能走的坐标点都进入队列
        {   sq[front].di ++;
            switch(sq[front].di)
            {
                case 0 :
                    i = sq[front].i - 1; j = sq[front].j ; if(mg2[i][j] == 0) find = 1; break;
                case 1 :
                    i = sq[front].i; j = sq[front].j + 1 ; if(mg2[i][j] == 0) find = 1; break;
                case 2 :
                    i = sq[front].i + 1; j = sq[front].j ; if(mg2[i][j] == 0) find = 1; break;
                case 3 :
                    i = sq[front].i; j = sq[front].j - 1; if(mg2[i][j] == 0) find = 1; break;
            }
            if(find == 1)        //此处根据find的值判断是否找到可走的元素，然后根据情况判断是否应该如队列
            {

                rear++;
                mg2[i][j] = -1;
                sq[rear].i = i; sq[rear].j = j; sq[rear].di = -1; sq[rear].pre = front;
            }
            find = 0;
        }
    }
}

//广度遍历
void main()
{
    mgpath2(1,1,8,8);
}