/*棧實現走迷宮
棧和遞迴思路有些不一樣，遞迴考慮一個結點的所有可能，不去觸碰下一個結點。
棧的話迴圈之前的初始條件要和下次迴圈一致，並且會用到提前判斷，提前繼續迴圈，
通過取棧頂達到假回溯效果，進到迴圈裡的結點雖然還是一樣，但是沒有記錄狀態了
遞迴的回溯狀態和之前是完全一樣的（在判斷條件的執行流中沒有出來），因此有試探-回溯的做法
*/
int passMaze(int maze[ROW][COL], Position entry, Stack* path) {
    //合法入口
    if (maze[entry.x][entry.y] != 1) {
        return path;
    }
    pushBack(path, entry);
    while
 
 (!isEmpty(path)) {
        Position cur = top(path);
        maze[cur.x][cur.y] = 2;
        if ((cur.x == ROW - 1 || cur.y == COL - 1 || cur.x == 0 || cur.y == 0)&& !(cur.x ==entry.x && cur.y ==entry.y)) {
            break;
        }
        else {
            int dir[4][2] = { { -1
 
,0 },{ 1 ,0 },{ 0 , -1 },{ 0, 1 } };
            int i;
            for (i = 0; i < 4; i++) {
                int xNext = cur.x + dir[i][0];
                int yNext = cur.y + dir[i][1];
                Position next;
                next.x = xNext, next.y = yNext;
                if (maze[next.x][next.y] == 1
 
) {
                    pushBack(path, next);
                    break;
                }
            }
            if (i != 4) {
                continue;
            }
            pop(path);
        }
    }
    return !isEmpty(path);
}
//遞迴走迷宮,單純的dfs是不能記錄有序路徑的
int passMazeR(int maze[ROW][COL], Position cur, Position entry, Stack* path)
{
    //不合法
    if (maze[cur.x][cur.y] != 1) {
        return 0;
    }
    //出口
    if ((cur.x == ROW - 1 || cur.y == COL - 1 || cur.x == 0 || cur.y == 0) 
        && !(cur.x == entry.x && cur.y == entry.y))
    {
        pushBack(path, cur);
        return 1;
    }
    maze[cur.x][cur.y] = 2;
    pushBack(path, cur);
    //四方
    int dir[4][2] = { { -1,0 },{ 1 ,0 },{ 0 , -1 },{ 0, 1 } };
    int i;
    for (i = 0; i < 4; i++) {
        int xNext = cur.x + dir[i][0];
        int yNext = cur.y + dir[i][1];
        Position next;
        next.x = xNext, next.y = yNext;
        if (passMazeR(maze, next, entry, path)) {
            return 1;
        }
    }
    pop(path);
    maze[cur.x][cur.y] = 1;
    return 0;
}