Question

我已经编写了一个简单的递归迷宫解决问题，该问题使用递归来找到最少要解决的动作。但是，程序死胡同，无法备份跟踪的路径。

为了解决该问题，我开始编写move函数的反函数。它们可以用来反转路径，但仍然需要某种方式来标识要使用的路径。

迷宫测试文件：

SWWWW
OOOOE
OWOWW
OOOWW

代码体：

//Read in maze
ifstream mazeFile;
mazeFile.open("MazeSample.txt");

vector<string> mazeRead{ istream_iterator<string>(mazeFile), istream_iterator<string>() };
maze = mazeRead;


    //Move checks
vector<string> maze;
int numMoves;
int leastMoves = 1000;

int row;
int column;

bool canMoveUp(int row, int column) {
    try {
        if (maze.at(row - 1).at(column) != ('O')) {
            cout << "(Can't move up)" << endl;
            if (maze.at(row - 1).at(column) == 'E') {
                return true;
            }
            return false;
        }
    }
    catch (const out_of_range& error) {
        cout << "(Can't move up)" << endl;
        return false;
    }
    return true;
}

bool canMoveDown(int row, int column) {
    try {
        if (maze.at(row + 1).at(column) != ('O')) {
            cout << "(Can't move down)" << endl;
            if (maze.at(row + 1).at(column) == 'E') {
                return true;
            }
            return false;
        }
    }
    catch (const out_of_range& error) {
        cout << "(Can't move down)" << endl;
        return false;
    }
    return true;
}

bool canMoveLeft(int row, int column) {
    try {
        if (maze.at(row).at(column - 1) != ('O')) {
            cout << "(Can't move left)" << endl;
            if (maze.at(row).at(column - 1) == 'E') {
                return true;
            }
            return false;
        }
    }
    catch (const out_of_range& error) {
        cout << "(Can't move left)" << endl;
        return false;
    }
    return true;
}

bool canMoveRight(int row, int column) {
    try {
        if (maze.at(row).at(column + 1) != ('O')) {
            cout << "(Can't move right)" << endl;
            if (maze.at(row).at(column + 1) == 'E') {
                return true;
            }
            return false;
        }
    }
    catch (const out_of_range& error) {
        cout << "(Can't move right)" << endl;
    }
    return true;
}


    //Maze solve function
void solve(int row, int column) {
    numMoves = numMoves + 1; //count moves

    //Base case (solution found; current position is 'E')
    if (maze[row][column] == 'E') {
        if (numMoves < leastMoves) {
            leastMoves = numMoves;
        }
    }

    if (maze[row][column] != 'E') {
        maze[row][column] = 't'; //mark path
    }

    // move up and see if move leads to solution (recursively)
    if (canMoveUp(row, column)) {
        cout << "(Move up)" << endl;
        row = row - 1;
        column = column;
        solve(row, column);
    }

    // if move chosen above doesn't lead to solution, move down & check
    if (canMoveDown(row, column)) {
        cout << "(Move down)" << endl;
        row = row + 1;
        column = column;
        solve(row, column);
    }

    // if move chosen above doesn't lead to solution, move left & check
    if (canMoveLeft(row, column)) {
        cout << "(Move left)" << endl;
        row = row;
        column = column - 1;
        solve(row, column);
    }

    // if move chosen above doesn't lead to solution, move right & check
    if (canMoveRight(row, column)) {
        cout << "(Move right)" << endl;
        row = row;
        column = column + 1;
        solve(row, column);
    }

    // if no above solution works, then unmark cell
    //backtrack (keeps going until all solutions reached)
    maze[row][column] = 'O';
    cout << "Mark as 'O'";
    numMoves = numMoves - 1;


    //TODO: PROBLEM: ROW/COLUMN NOT RESET AFTER STUCK; KEEPS SAME VALUE
            //Questionable code
    if (!canMoveUp(row, column)) {
        //Inverse of canMove?
        row = row + 1;
        column = column;
    }


    //Display vector contents
    cout << endl;
    for (int row = 0; row < maze.size(); row++) {
        cout << endl;
        for (int column = 0; column < maze[row].size(); column++) {
            cout << maze[row][column];
        }
    }
    cout << endl;
}

当走到尽头时，我希望迷宫能够回到带有运动选项的最后一个交界处。而是，光标来回移动，无法解决。这可能是由于执行了移动功能所致；如果能够移动，它将行/列变量设置为新的空格。

错误路径如下所示，在第1行第1列的't'和'O'之间切换：

SWWWW
tttOE
tWtWW
tttWW

SWWWW
tOtOE
tWtWW
tttWW

一旦无法在四个方向中的任何一个上移动，我希望代码撤消之前的移动，直到到达最后一个结点。

Answer 1

无需深入了解算法

if (canMoveUp(row, column)) {
    cout << "(Move up)" << endl;
    row = row - 1;
    column = column;
    solve(row, column);
}

看起来很腥。您正在更改在后续块中使用的变量row和column。如果canMoveUp为true，则下一个检查将为canMoveDown('originalrow' - 1, column)，这将失败（因为它再次向该行添加1并检查刚刚标记为t的字段。

不是吗

if (canMoveUp(row, column)) {
    cout << "(Move up)" << endl;
    solve(row - 1, column);
}

使用迷宫求解程序回溯逻辑错误

1 个答案: