数据结构 C语言 双向链栈 迷宫问题

【问题描述】
以一个mXn的矩阵表示迷宫，0和1分别表示迷宫中的通路和障碍。设计一个程序，对任意设定的迷宫，求出一条从入口到出口的通路，或得出没有通路的结论。
【任务要求】
实现链栈求解迷宫从入口到出口的一条可行通路。
【测试数据】
迷宫的测试数据如下：左上角（0，1）为入口，右下角（8，9）为出口。

#include<stdio.h>#include<stdlib.h>int count = 0;typedef struct {    int x, y;}position;typedef struct node{    position data;    struct node *next, *pre;}linkNode;typedef struct {    linkNode *head;    linkNode *top;}linkStack;int initStack(linkStack *s){    s->head = (linkNode*)malloc(sizeof(linkNode));    s->top = s->head;    s->top->next = NULL;    s->top->pre = NULL;}int pushStack(linkStack *s, position value){    linkNode *p;    p = (linkNode*)malloc(sizeof(linkNode));    s->top->data = value;    p->next = s->top->next;    p->pre = s->top;    s->top->next = p;    s->top = p;    count++;    return 0;}int popStack(linkStack *s, position *nowPos){    s->top = s->top->pre;    free(s->top->next);    count--;    *nowPos = s->top->pre->data;}int main(){    linkStack s;    linkNode *p;    position nowPos, startPos, endPos;    int i, m, n, j;    int **map;    initStack(&s);    printf("请输入行列数\n");    scanf("%d%d", &m, &n);    map = (int**)malloc(sizeof(int)*m);    for (i = 0; i < m; i++) {        map[i] = (int*)malloc(sizeof(int)*n);    }    printf("请输入地图:\n");    for (i = 0; i < m; i++) {        for (j = 0; j < n; j++) {            scanf("%d", &map[i][j]);        }    }    printf("\n");    printf("请输入开始位置和结束位置:\n");    scanf("%d %d", &startPos.x, &startPos.y);    scanf("%d %d", &endPos.x, &endPos.y);    nowPos = startPos;    pushStack(&s, nowPos);    while ((nowPos.x != endPos.x) || (nowPos.y != endPos.y)) {        if (0 == map[nowPos.x + 1][nowPos.y]) {            nowPos.x += 1;            map[nowPos.x][nowPos.y] = 2;            pushStack(&s, nowPos);        }        else if (0 == map[nowPos.x][nowPos.y - 1]) {            nowPos.y -= 1;            map[nowPos.x][nowPos.y] = 2;            pushStack(&s, nowPos);        }        else if (0 == map[nowPos.x - 1][nowPos.y]) {            nowPos.x -= 1;            map[nowPos.x][nowPos.y] = 2;            pushStack(&s, nowPos);        }        else if (0 == map[nowPos.x][nowPos.y + 1]) {            nowPos.y += 1;            map[nowPos.x][nowPos.y] = 2;            pushStack(&s, nowPos);        }        else {            map[nowPos.x][nowPos.y] = 2;            popStack(&s, &nowPos);        }    }    printf("路径为:\n");    for (i = 0, p = s.head; i < count; i++,p = p->next) {        printf("%d,%d\n", p->data.x, p->data.y);    }    system("pause");    return 0;}