数据结构（C语言） 栈的应用之迷宫求解

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
typedef struct Pos {
    int x;
    int y;
}Pos;
typedef struct SqStack {
    Pos data[200];
    int top;
}SqStack;
void Push(SqStack* S, Pos e) {
    S->top++;
    S->data[S->top].x = e.x;
    S->data[S->top].y = e.y;

}
void Pop(SqStack* S, Pos* e) {
    e->x = S->data[S->top].x;
    e->y = S->data[S->top].y;
    S->top--;
}
int Empty(SqStack* S) {
    return S->top == -1;
}
void Init(SqStack* S) {
    S->top = -1;
}
//maze[i][j] 1表示障碍，0表示通路,返回路径
SqStack* MazePath(int maze[10][10], Pos start, Pos end) {
    int m[10][10];
    memcpy(m, maze, 10 * 10 * sizeof(int));
    SqStack* S = (SqStack*)malloc(sizeof(SqStack));
    Init(S);
    Pos curpos = start;
    if (m[curpos.y-1][curpos.x-1] == 0) {
        m[curpos.y - 1][curpos.x - 1] = 3;
        Push(S, curpos);
        if (curpos.x == end.x&&curpos.y == end.y)
            return S;
        else
            curpos.x++;
    }
    else {
        return NULL;
    }
    while (!Empty(S)) {
        if (m[curpos.y - 1][curpos.x - 1] == 0) {
            m[curpos.y - 1][curpos.x - 1] = 3;
            Push(S, curpos);
            if (curpos.x == end.x&&curpos.y == end.y)
                return S;
            else
                curpos.x++;
        }
        else {
            Pos prepos;
            Pop(S, &prepos);
            if (curpos.x > prepos.x) {
                curpos.x = prepos.x;
                curpos.y = prepos.y + 1;
                Push(S, prepos);
            }
            else if (curpos.y > prepos.y) {
                curpos.x = prepos.x - 1;
                curpos.y = prepos.y;
                Push(S, prepos);
            }
            else if (curpos.x < prepos.x) {
                curpos.x = prepos.x;
                curpos.y = prepos.y - 1;
                Push(S, prepos);
            }
            else {
                m[prepos.y - 1][prepos.x - 1] = 2;
                curpos = prepos;
            }
        }
    }
    return NULL;
}

void main() {
    int maze[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, 0, 0, 0, 0, 0, 0, 0, 0, 1
    };
    Pos start = { 2,2 };
    Pos end = { 9,9 };
    SqStack* S = MazePath(maze, start, end);
    SqStack path;
    Init(&path);
    if (S) {
        while (!Empty(S)) {
            Pos p;
            Pop(S, &p);
            Push(&path, p);
        }
        free(S);
    }
    else {
        printf("不存在路径");
    }
    while (!Empty(&path)) {
        Pos p;
        Pop(&path, &p);
        printf("(%d,%d)\n", p.x,p.y);
    }
    getchar();
}