队列的其本应用_迷官问题

问题：
利用队列的假出队，实现迷官问题寻解。
要求：
0 墙 1 路， 寻找最短路， 如果有多条，输出其中一条路。

main.c

#include <stdio.h>
#include <stdlib.h>

#define M 5
int map[M+2][M+2];
int front = 0, rear = 0;
typedef struct Point_def{
    int x;
    int y;
    int pre;
}Point;
Point q[(M+2)*(M+2)];
typedef struct direction_def{
    int x;
    int y;
}Direction;
Direction direct[4] = {
    {-1,0}, {0,1}, {1,0}, {0, -1}
};
int isEmpty(){
    return front==rear?1:0;
}
void initMap(){
    int i, j;
    for (i=0; i<M+2; i++)
        for (j=0; j<M+2; j++){
            if (i == 0 || i == M+1 || j == 0 || j == M+1){
                map[i][j] = 0; continue;
            }
            scanf("%d", &map[i][j]);
        }
}
void enQueue(Point p){
    q[rear++] = p;
}
Point deQueue(){
    if (isEmpty()) 
        exit(1);
    return q[front++];
}

void findPath(int start_x, int start_y, int end_x, int end_y){
    int flag = 0;
    int d;
    int i, j;
    Point tmp_p1={start_x, start_y, -1};
    Point tmp_p2;
    enQueue(tmp_p1);
    while (!flag){
        tmp_p1 = deQueue();
        map[tmp_p1.x][tmp_p1.y] = -1;
        for (d=0; d<4; d++){
            i = tmp_p1.x + direct[d].x;
            j = tmp_p1.y + direct[d].y;
            if (map[i][j] == 1){
                tmp_p2.x = i; tmp_p2.y = j; tmp_p2.pre = front-1;
                enQueue(tmp_p2);
            }
            if (i == end_x && j == end_y){
                tmp_p2.x = i; tmp_p2.y = j; tmp_p2.pre = front-1;
                enQueue(tmp_p2);
                flag = 1;  
                break;
            }

        }// end for 
    }// end while 
}// end findPath

void printPath(){
    int k = rear-1;
    while (k != -1){
        printf("(%d,%d)
", q[k].x, q[k].y);
        k = q[k].pre;
    }
}

int main(){
    int flag = 0;
    int start_x = 1, start_y = 1;
    int end_x = 5, end_y = 5;
    initMap();
    findPath(start_x, start_y, end_x, end_y);
    printPath();
    return 0;
}