如何找到所有可用的迷宫路径？

Question

我正在尝试编写一个程序，该程序被赋予一个迷宫并试图找到出路。 M是入口，E是出口，1是墙壁，0是路径。它应该找到每条路径并将P放入该路径中。它应该找到所有可用的路径。现在它找到了路径的一部分。

这是代码：

public class Maze 
{
    private int size;
    private String[][] board;
    private int total; //# of boards
    private int eX;
    private int eY;
    private int mX;
    private int mY;

    public Maze( int size, String[][] board )
    {
        this.size = size;
        this.board = board;
        total = 0;

    }

    private void find( String c )
    {
        int x=0, y=0;
        for( int i = 0; i < size; i++ )
        {
            for( int j = 0; j < size; j++ )
            {
                if( board[i][j].equals(c) )
                {
                    x = i;
                    y = j;
                }
            }
        }
        if( c.equals("M") )
        {
            mX = x;
            mY = y;
        }
        else if( c.equals("E") )
        {
            eX = x;
            eY = y;
        }
    }

    public void findPath(  )
    {
        find( "M" );
        find( "E" );
        findNext( mX, mY );
    }

    public void findNext( int x, int y )
    { 
        String last = board[x][y];
            if( board[x][y].equals("P") ) board[x][y] = "1";
        board[x][y] = "P";

        if( rightAvailability(x,y) )
        {
            findNext(x+1, y);
        }
        else if( leftAvailability(x,y) )
        {
            findNext(x-1, y);
        }
        else if( aboveAvailability(x,y) )
        {
            findNext(x, y+1);
        }
        else if( belowAvailability(x,y) )
        {
            findNext(x, y-1);
        }
        else
        {
            total++;
            printBoard();
        }

        board[x][y]= last;
    }

    public boolean rightAvailability( int x, int y )
    {
        if( x+1 >= size ) return false;
        else if( board[x+1][y].equals("1") ) return false;
        else if( board[x+1][y].equals("P") ) return false;
        else return true;
    }
    public boolean leftAvailability( int x, int y )
    {
        if( x-1 < 0) return false;
        else if( board[x-1][y].equals("1") ) return false;
        else if( board[x-1][y].equals("P") ) return false;
        else return true;
    }
    public boolean aboveAvailability( int x, int y )
    {
        if( y+1 >= size ) return false;
        else if( board[x][y+1].equals("1") ) return false;
        else if( board[x][y+1].equals("P") ) return false;
        else return true;
    }
    public boolean belowAvailability( int x, int y )
    {
        if( y-1 < 0) return false;
        else if( board[x][y-1].equals("1") ) return false;
        else if( board[x][y-1].equals("P") ) return false;
        else return true;
    }

    public void printBoard()
    {
        System.out.println( "The board number " +total+ " is: ");
        for(int i=0; i< size; i++ )
        {
            for(int j=0; j< size; j++ )
            {
                if( (i==mX) && (j==mY) )
                {
                    System.out.print("M");
                }
                else if( (i==eX) && (j==eY) )
                {
                    System.out.print("E");
                }
                else if( board[i][j].equals("1") )
                {
                    System.out.print("1");
                }
                else if( board[i][j].equals("0") )
                {
                    System.out.print("0");
                }
                else
                {
                    System.out.print("P");
                }
            }
            System.out.println();
        }
    }
}

这是测试器：

public class MazeTester 
{
    public static void main(String[] args) 
    {
        int size = 11;
        String[][] board = new String[][]
            {
                {"1","1","1","1","1","1","1","1","1","1","1"},
                {"1","0","0","0","0","0","1","0","0","0","1"},
                {"1","0","1","0","0","0","1","0","1","0","1"},
                {"E","0","1","0","0","0","0","0","1","0","1"},
                {"1","0","1","1","1","1","1","0","1","0","1"},
                {"1","0","1","0","1","0","0","0","1","0","1"},      
                {"1","0","0","0","1","0","1","0","0","0","1"},
                {"1","1","1","1","1","0","1","0","0","0","1"},
                {"1","0","1","M","1","0","1","0","0","0","1"},
                {"1","0","0","0","0","0","1","0","0","0","1"},
                {"1","1","1","1","1","1","1","1","1","1","1"},  
            };


        Maze m = new Maze( size, board );
        m.findPath();
    }
}

这是当前输出：

The board number 1 is: 
11111111111
1PPPPP1PPP1
1P1PPP1P1P1
EP1PPPPP1P1
101111101P1
10101PPP1P1
10001P1PPP1
11111P1PP01
101M1P1PP01
100PPP1PP01
11111111111
The board number 2 is: 
11111111111
1PPPPP1PPP1
1P1PPP1P1P1
EP100PPP1P1
101111101P1
10101PPP1P1
10001P1PPP1
11111P1PP01
101M1P1PP01
100PPP1PP01
11111111111
The board number 348 is: 
11111111111
1PPPPP10001
1P1PPP10101
EP1PPPPP101
1011111P101
10101PPP101
10001P10001
11111P10001
101M1P10001
100PPP10001
11111111111

编辑：我添加了 if( board[x][y].equals("P") ) board[x][y] = "1";在 findIndex 的开头。我还解决了 x <= 0 问题。我将输出更新为我现在得到的（实际上是打印 348 个类似的板）。

Answer 1

我在行中放置了部分猜测：

else if( belowAvailability(x,y) )
{
        findNext(x, x-1);
}

x-1 应该是 y-1

您会发现的另一个问题是您正在使用 else if 块。如果你遇到一个分支，说

1E1
101
1P0
1P1

你会先尝试向右走，然后当它严重失败时，它就不会尝试向上。您实际上可以在测试输出中看到，

_._._789_._
_..._6_A54_
_____5_B23_
_._M_4_C10_
_..123_DEF_
___________

为了方便阅读，以十六进制编号。它进入右下角，然后卡住；由于无处可去，棋盘被打印出来，递归结束时不会回溯到未经测试的方块。

再次编辑。仍在寻找，但在左/右可用性中，您有另一个 x/y 不匹配，并且您可能只想在 x-1 < 时拒绝可用性。 0（或y-1）；因为目标节点位于 y=0。

通过这些更改，并且仅当 x==eX && 时才触发打印y==eY，我正在得到解决方案的输出。解决方案有很多，但是解决方案。

关于编辑计数的另一个幽默事实是：您可以向后/向左和向上/向下。 x 坐标指定您正在查看的输入的行，y 坐标指定列。在这种情况下使用 r/c 是相当常见的。

Answer 2

标准路径查找算法应该可以工作，您需要修改它们以匹配您的世界定义。

但 A* 或 D* 算法效果很好。他们使用图表，您应该能够根据您的世界定义来定义该图表。 (http://en.wikipedia.org/wiki/A*)

Dijstra 算法也应该努力寻找路径（再次使用图表）。它通常用于网络路由 - 但它也适用于正常的路径查找。 (http://en.wikipedia.org/wiki/Dijkstra%27s_algorithm)

基本上我的方法是将迷宫定义转换为图表（节点是“连接点”，边缘是“走廊”），然后使用其中一种算法。