如何找到所有可用的迷宫路径？

StackOverflow https://stackoverflow.com/questions/1523323

  •  19-09-2019
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题

我想写被赋予一个迷宫，并试图找到出路的程序。 M是入口，E是出口，1秒的墙壁和0的途径。它应该找到每一个路径和放P的路径。它应该找到的所有可用路径。现在它找到一个路径的一部分。

下面是代码：

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();
        }
    }
}

下面是测试器：

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();
    }
}

下面是电流输出：

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：我添加如果（板[X] [Y] .equals（ “P”））板[X] [Y] = “1”;在findIndex的开始。我还固定在x <= 0的问题。我更新了输出什么，我现在得到（它实际上是打印348块类似的板）。

有帮助吗？

解决方案

我把一个局部猜测的行：

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

X-1应该是Y-1

你会发现另一个问题是事实，你正在使用否则，如果块。如果你遇到一个分支，比方说

1E1
101
1P0
1P1

您会尽量向右走，然后再当失败草草收场，也不会尝试去了。实际上，你可以看到，在测试输出，

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

在十六进制编号为方便阅读的方便。它进入右下角，然后卡;具有无处可去的板印刷，并递归无回溯到未经测试的正方形结束。

再次编辑。还在寻找，但在左/右可用性您有其他的x / y不匹配，你可能只是想否认可用性当x-1 <0（或y-1）;为目标节点是在y = 0。

使用这些变化，并且具有打印触发仅当x ==的eX &&ÿ== EY，我发现了解决方案的输出。一个伟大的很多解决方案，但解决方案。

有关编辑算一个比较幽默的事实：你左/右和上/下倒退。 x坐标上，指定您希望在输入的行，而y坐标指定列。它相当常见的是使用R / C在这样的情况下。

其他提示

标准寻路算法应该工作，你需要对其进行修改以符合你的世界定义。

但是，A *或d *算法工作得很好。他们使用的图形，你应该能够从你的世界定义定义。 （ http://en.wikipedia.org/wiki/A *）

另外Dijstra算法应当在发现一个路径（再次使用的曲线图）工作。它通常用于网络路由 - 但它也适用于正常路径的发现。 （ http://en.wikipedia.org/wiki/Dijkstra%27s_algorithm ）

基本上我的方法是把你的迷宫定义成图形（节点被“连接点”，边缘是“走廊”），然后使用这些算法中的一个。

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