通过回溯在迷宫中寻找路径

Question

我想编写一个程序来通过回溯查找迷宫中从右上角到左下角的路径是否存在。输入数字是 n 和 m，它们是矩形迷宫和迷宫的维度，字符 '.'表示您可以通过的图块，字符“x”表示您无法通过的图块。我已经编写了代码，它相当简单，但什么都没有显示，而它应该显示“da”（塞尔维亚语“yes”）和“ne”（塞尔维亚语“no”）。

#include <bits/stdc++.h>

using namespace std;

bool maze[20][20]; //defined maze of maximum size 20x20

//checking if a position is viable for moving through
bool Safe(int n, int m, int x, int y)
{
    if(x >= 0 && x < n && y >= 0 && y < m)
    {
        if(maze[x][y] == 1) return true;
    }
    return false;
}

bool Utility(int n, int m, int x, int y) //main utility function
{
    if(x == n - 1 && y == m - 1 && maze[x][y] == 1) // base case, end of maze
    {
        return true;
    }

    if(Safe(n, m, x, y))
    {
        if(Safe(n, m, x + 1, y)) // checking if it is viable to move down
        {
            if(Utility(n, m, x + 1, y))
            {
                return true;
            }
        }
        if(Safe(n, m, x, y + 1))
        {
            if(Utility(n, m, x, y + 1)) // checking if it is viable to move right
            {
                return true;
            }
        }
        if(Safe(n, m, x - 1, y))
        {
            if(Utility(n, m, x - 1, y)) // checking if it is viable to move up
            {
                return true;
            }
        }
        if(Safe(n, m, x, y - 1))
        {
            if(Utility(n, m, x, y - 1)) // checking if it is viable to move left
            {
                return true;
            }
        }
    }

    return false; // returning false
}

int main()
{
    int n, m;

    cin >> n >> m; // input dimensions of the maze

    for(int i = 0; i < n; i++) // input maze
    {
        for(int j = 0; j < m; j++)
        {
            char c;
            cin >> c;
            if(c == '.') //character '.' means a tile which you can go through
            {
                maze[i][j] = 1;
            }
            else         //character 'x' means a tile which you cannot go through
            {
                maze[i][j] = 0;
            }
        }
    }

    if(Utility(n, m, 0, 0)) //printing yes or no
    {
        cout << "da";
    }
    else
    {
        cout << "ne";
    }
    

    return 0;
}

示例输入：

8 8 
.x.....x 
.x.x.x.x 
.x.x.x.x 
.x.x.x.x 
.x.x.x.x 
.x.x.x.x 
.x.x.x.x 
...x.x..

示例输出：da

Answer 1

问题是，假设你从(0, 0) -> (1, 0) 出发，然后在(1, 0) 你可以再次回到(0, 0)，这将永远循环。为避免这种情况，我创建了一个 visited 数组，如果单元格 (x, y) 已被访问，则其值为 true，否则为 false。

我已经用///////////// change here /////////////评论标记了我在哪里进行了更改

#include <bits/stdc++.h>

using namespace std;

bool maze[20][20]; //defined maze of maximum size 20x20
///////////// change here /////////////
bool visited[20][20];

bool Safe(int n, int m, int x, int y) //checking if a position is viable for moving through
{
    if(x >= 0 && x < n && y >= 0 && y < m)
    {
        if(maze[x][y] == 1) return true;
    }
    return false;
}

bool Utility(int n, int m, int x, int y) //main utility function
{
    if(x == n - 1 && y == m - 1 && maze[x][y] == 1) // base case, end of maze
    {
        return true;
    }

    ///////////// change here ///////////// 
    if(!visited[x][y] && Safe(n, m, x, y))
    {
        ///////////// change here /////////////
        visited[x][y] = true;

        if(Safe(n, m, x + 1, y)) // checking if it is viable to move down
        {
            if(Utility(n, m, x + 1, y))
            {
                return true;
            }
        }
        if(Safe(n, m, x, y + 1))
        {
            if(Utility(n, m, x, y + 1)) // checking if it is viable to move right
            {
                return true;
            }
        }
        if(Safe(n, m, x - 1, y))
        {
            if(Utility(n, m, x - 1, y)) // checking if it is viable to move up
            {
                return true;
            }
        }
        if(Safe(n, m, x, y - 1))
        {
            if(Utility(n, m, x, y - 1)) // checking if it is viable to move left
            {
                return true;
            }
        }
    }

    return false; // returning false
}

int main()
{
    int n, m;

    cin >> n >> m; // input dimensions of the maze

    for(int i = 0; i < n; i++) // input maze
    {
        for(int j = 0; j < m; j++)
        {
            char c;
            cin >> c;
            if(c == '.') //character '.' means a tile which you can go through
            {
                maze[i][j] = true;
            }
            else         //character 'x' means a tile which you cannot go through
            {
                maze[i][j] = false;
            }
            ///////////// change here /////////////
            visited[i][j] = false;
        }
    }

    if(Utility(n, m, 0, 0)) //printing yes or no
    {
        cout << "da";
    }
    else
    {
        cout << "ne";
    }
    

    return 0;
}

这是我测试它的链接：https://ideone.com/vVqAjF