数据结构 C语言双向链栈迷宫问题

柔光的暖阳◎ 2022-06-13 09:22 227阅读 0赞

【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。![矩阵模拟地图][SouthEast]  
![迷宫地图][SouthEast 1]

#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;
    }

[SouthEast]: /images/20220613/32474971c1e34c479bba617fcccfe8a1.png
[SouthEast 1]: /images/20220613/2b70755f65944a6680393e846ca3ca24.png