c++ - 使用BFS的迷宫路线输出错误

Question

我正在创建一个程序，它将通过迷宫找到最短的路线，并将路线输出为字符串。我已经利用BFS查找最短的移动量，现在我需要输出这些移动量是什么。在我的代码中，我尝试使用switch语句在解决迷宫问题时向字符串添加字符。它利用了for循环中的值，但这没有成功。我还尝试创建2个不同的开关，其中一个将检查行int，另一个将检查col int。但是，这也不成功。
我可以解决一些难题，例如:

xxxxx
x...B
A...x
xxxxx

正确输出:

Shortest route is: ENEEE
Shortest Path is 5

但是，我要解决的另一个迷宫是:

xxxxxxxxxx
x........x
x..xxxx..x
x..xxxx..x
A..xxxx..B
x..xxxx..x
x........x
x........x
xxxxxxxxxx

结果是:

Shortest route is: ENESNESSNESSE
Shortest Path is 13

虽然距离正确，但路线本身却不正确。如果有人能发现我想念的东西，请告诉我。
请在下面找到我的代码:

#include <iostream>
#include <queue>
#include <vector>
#include <fstream>
#include <string>


using namespace std;


struct Point
{
    int x;
    int y;
};

struct queueNode
{
    Point pt;  
    int dist;  
};


bool isValid(int row, int col, int x, int y)
{

    return (row >= 0) && (row < x) && (col >= 0) && (col < y);
}

bool isSafe(vector<vector<char>> maze, vector<vector<int>> visited, int x, int y)
{
    if (maze[x][y] == 'x' || visited[x][y] || maze[x][y] == NULL)
        return false;
    else
    return true;
}

int rowNum[] = { -1, 0, 0, 1 };
int colNum[] = { 0, -1, 1, 0 };

int solveMaze(vector<vector<char>> maze, Point src, Point dest, int rows, int columns)
{
    string route;

    if (!maze[src.x][src.y] || !maze[dest.x][dest.y])
        return -1;

    auto visited = vector<vector<int>>(rows, vector<int>(columns));

    // Mark the source cell as visited
    visited[src.x][src.y] = true;

    // Create a queue for solveMaze
    queue<queueNode> q;

    // Distance of source cell is 0
    queueNode s = { src, 0 };
    q.push(s);  // Enqueue source cell

    // Do a solveMaze starting from source cell
    while (!q.empty())
    {
        queueNode curr = q.front();
        Point pt = curr.pt;

        // If we have reached the destination cell,
        // we are done
        if (pt.x == dest.x && pt.y == dest.y)
        {
            route.resize(curr.dist);
            cout << "Shortest route is: " << route << endl;
            return curr.dist;
        }

        // Otherwise dequeue the front 
        // cell in the queue
        // and enqueue its adjacent cells
        q.pop();

        for (int i = 0; i < 4; i++)
        {
            int row = pt.x + rowNum[i];
            int col = pt.y + colNum[i];

            // if adjacent cell is valid, has path and
            // not visited yet, enqueue it.
            if (isValid(row, col, rows, columns) && maze[row][col] && !visited[row][col] && isSafe(maze,visited,row,col))
            {
                switch (i)
                {
                case 0:
                    route.append("N");
                    break;
                case 1:
                    route.append("W");
                    break;
                case 2:
                    route.append("E");
                    break;
                case 3:
                    route.append("S");
                    break;
                }
                visited[row][col] = true;
                queueNode Adjcell = { {row, col},curr.dist + 1 };
                q.push(Adjcell);
            }
        }
    }

    return -1;
}

int main()
{
    string filename;
    int rows;
    int columns;
    cout << "Please input the name of your maze!" << endl;
    cin >> filename;
    cout << endl;
    cout << "Please input the amount of rows in your maze!" << endl;
    cin >> rows;
    cout << endl;
    cout << "Please input the amount of columns in your maze!" << endl;
    cin >> columns;
    cout << endl;
    int startRow = 0;
    int startColumn = 0;
    int endRow = 0;
    int endColumn = 0;
    int column = 0;
    auto maze = vector<vector<char>>(rows,vector<char>(columns));    
    ifstream input(filename);
    char data = input.get();
    while (!input.eof())
    {
        for (int row = 0; row < rows; row++)
        {
            while (data != '\n' && !input.eof())
            {
                if (data == 'A')
                {
                    startRow = row;
                    startColumn = column;
                }
                if (data == 'B')
                {
                    endRow = row;
                    endColumn = column;
                }
                maze[row][column] = data;
                column++;
                data = input.get();
            }
            column = 0;

            data = input.get();
        }
    }
    input.close();
    cout << "The Maze being solved is: " << endl;
    for (int y = 0; y < rows; y++)
    {
        for (int x = 0; x < columns; x++)
        {
            cout << maze[y][x];
        }
        cout << endl;
    }
    Point source = { startRow, startColumn };
    Point dest = { endRow, endColumn };

    int dist = solveMaze(maze, source, dest,rows,columns);

    if (dist != -1)
        cout << "Shortest Path is " << dist;
    else
        cout << "Shortest Path doesn't exist";

    return 0;

}

Answer 1

在我看来，您在BFS期间将char附加到路由中。这将不会返回正确的最短路径，因为代码需要验证哪个方向实际上与最短路径相对应。通过在实际测试路径的最短路径部分之前添加路径的方向，输出实际上仅告诉我们BFS本身的执行路径。
为了解决这个问题，我们使用一个父数组，该数组保留每个节点的父级(当您第一次访问相邻单元格时，将该单元格的父级设置为当前单元格)。广度优先搜索完成后，将“tracer”变量设置为目标，并使用父数组向后追溯，直到到达起点为止。请注意，最后的“路径重构”以相反的顺序重构路径(因为我们从目的地迭代回到起点)。
这是我实现此方法的代码:

#include <iostream>
#include <fstream>
#include <cmath>
#include <algorithm>
#include <vector>
#include <queue>
#include <map>

using namespace std;

void bfs(vector <vector<char>> &grid, int R, int C, int sR, int sC, int eR, int eC){
    vector <vector<bool>> visited(R, vector <bool> (C, false));
    map <pair<int, int>, pair <int, int>> parent;
    queue <pair<int, int>> Q;

    Q.push({sR, sC});
    visited[sR][sC] = true;
    parent[{sR, sC}] = {sR, sC}; // it doesn't really matter what you set it to

    while(!Q.empty()){
        int r = Q.front().first, c = Q.front().second;
        Q.pop();

        if(r == eR && c == eC){
            break;
        }

        // enqueue all "valid" (unvisited, not-a-wall) adjacent squares
        if(r > 0 && !visited[r-1][c] && grid[r-1][c] != 'x'){
            visited[r-1][c] = true;
            parent[{r-1, c}] = {r, c};
            Q.push({r-1, c});
        }

        if(r < R-1 && !visited[r+1][c] && grid[r+1][c] != 'x'){
            visited[r+1][c] = true;
            parent[{r+1, c}] = {r, c};
            Q.push({r+1, c});
        }

        if(c > 0 && !visited[r][c-1] && grid[r][c-1] != 'x'){
            visited[r][c-1] = true;
            parent[{r, c-1}] = {r, c};
            Q.push({r, c-1});
        }

        if(c < C-1 && !visited[r][c+1] && grid[r][c+1] != 'x'){
            visited[r][c+1] = true;
            parent[{r, c+1}] = {r, c};
            Q.push({r, c+1});
        }
    }

    if(visited[eR][eC] == false){
        cout << "The destination was unreachable from the source." << endl;
        return;
    }

    // trace back from the destination to the start
    pair <int, int> tracer = {eR, eC};
    vector <char> path;

    // note that this loop constructs the path in reverse order because
    // we iterate from the end back to the start
    while(tracer != make_pair(sR, sC)){
        pair <int, int> next = parent[tracer];

        if(next.first - tracer.first == 1){
            path.push_back('N');
        }
        else if(next.first - tracer.first == -1){
            path.push_back('S');
        }
        else if(next.second - tracer.second == 1){
            path.push_back('L');
        }
        else if(next.second - tracer.second == -1){
            path.push_back('R');
        }

        tracer = next;
    }

    cout << "Shortest Path Length: " << path.size() << endl;

    // we print this in reverse order (see the above comment)
    for(int i = path.size()-1; i >= 0; --i){
        cout << path[i];
    }
    cout << endl;
}

int main()
{
    int R, C, sR, sC, eR, eC;
    cin >> R >> C;

    vector <vector<char>> grid(R, vector <char> (C));
    for(int i = 0; i < R; ++i){
        for(int j = 0; j < C; ++j){
            cin >> grid[i][j];

            if(grid[i][j] == 'A'){
                sR = i;
                sC = j;
            }
            else if(grid[i][j] == 'B'){
                eR = i;
                eC = j;
            }
        }
    }
    // Input Done

    bfs(grid, R, C, sR, sC, eR, eC);

    return 0;
}