递归迷宫求解器问题java

时间:2014-03-06 22:56:25

标签: java recursion solver maze

我现在已经让它停止无限重复,但它只是一遍又一遍地尝试相同的错误路径。有没有人知道如何让它尝试不同的路径?

数字的关键: 0是开放的 1是墙 2是路径的一部分 3是迷宫的终点

    public class Maze{
  public static void main(String[] args){
    int[][] maze = {{1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1},
      {0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,1},
      {1,0,1,1,1,1,1,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,1,0,1,1,1,0,1,1,1,1,1,1,1,0,1,1,1,0,1,0,1},
      {1,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,1,0,1,0,1},
      {1,0,1,0,1,1,1,0,1,0,1,1,1,1,1,1,1,1,1,0,1,0,1,1,1,0,1,0,1,1,1,1,1,0,1,1,1,0,1,0,1,0,1,1,1,0,1,1,1,0,1},
      {1,0,1,0,1,0,1,0,1,0,1,0,0,0,1,0,0,0,1,0,0,0,1,0,1,0,1,0,1,0,0,0,0,0,1,0,0,0,1,0,1,0,0,0,0,0,1,0,0,0,1},
      {1,0,1,1,1,0,1,0,1,1,1,0,1,0,1,0,1,0,1,1,1,1,1,0,1,0,1,0,1,1,1,0,1,1,1,0,1,1,1,0,1,1,1,1,1,1,1,0,1,0,1},
      {1,0,0,0,0,0,1,0,0,0,0,0,1,0,1,0,1,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,1,0,0,0,1,0,0,0,1,0,0,0,0,0,0,0,1,0,1},
      {1,1,1,0,1,1,1,1,1,1,1,1,1,0,1,0,1,0,1,0,1,0,1,1,1,0,1,1,1,1,1,1,1,0,1,0,1,0,1,1,1,0,1,1,1,1,1,1,1,0,1},
      {1,0,1,0,1,0,0,0,0,0,0,0,1,0,1,0,1,0,1,0,1,0,0,0,1,0,0,0,0,0,0,0,0,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,1},
      {1,0,1,0,1,0,1,1,1,1,1,0,1,0,1,0,1,1,1,0,1,1,1,0,1,1,1,1,1,1,1,0,1,1,1,1,1,1,1,0,1,0,1,1,1,0,1,0,1,1,1},
      {1,0,1,0,1,0,0,0,1,0,1,0,1,0,1,0,0,0,0,0,1,0,0,0,1,0,0,0,1,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,1},
      {1,0,1,0,1,1,1,0,1,0,1,0,1,0,1,1,1,1,1,1,1,0,1,0,1,0,1,1,1,0,1,1,1,0,1,1,1,1,1,1,1,1,1,0,1,1,1,1,1,1,1},
      {1,0,1,0,1,0,1,0,1,0,0,0,1,0,1,0,0,0,0,0,0,0,1,0,0,0,1,0,0,0,1,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,1},
      {1,0,1,0,1,0,1,0,1,0,1,1,1,0,1,0,1,1,1,1,1,1,1,1,1,0,1,0,1,1,1,0,1,1,1,0,1,1,1,1,1,0,1,0,1,1,1,1,1,0,1},
      {1,0,0,0,1,0,1,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,1,0,0,0,1,0,0,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,1,0,0,0,0,0,1},
      {1,0,1,1,1,0,1,0,1,1,1,1,1,1,1,0,1,0,1,1,1,1,1,0,1,1,1,1,1,0,1,0,1,0,1,1,1,0,1,0,1,1,1,1,1,0,1,1,1,0,1},
      {1,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,0,0,0,0,1,0,1,0,0,0,1,0,0,0,1,0,1,0,0,0,0,0,1,0,1,0,1},
      {1,1,1,1,1,0,1,1,1,0,1,1,1,0,1,0,1,0,1,0,1,1,1,1,1,1,1,0,1,0,1,1,1,1,1,0,1,0,1,0,1,0,1,1,1,1,1,0,1,0,1},
      {1,0,0,0,1,0,0,0,1,0,0,0,1,0,1,0,1,0,0,0,1,0,0,0,1,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,1,0,0,0,1,0,0,0,1,0,1},
      {1,0,1,1,1,1,1,0,1,1,1,0,1,1,1,0,1,0,1,1,1,0,1,0,1,1,1,0,1,1,1,1,1,0,1,0,1,1,1,1,1,1,1,0,1,0,1,0,1,0,1},
      {1,0,0,0,0,0,1,0,0,0,1,0,0,0,1,0,1,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,1,0,0,0,1,0,1,0,1,0,1,0,1},
      {1,1,1,0,1,0,1,1,1,0,1,1,1,0,1,0,1,1,1,1,1,0,1,1,1,0,1,1,1,0,1,0,1,0,1,1,1,0,1,0,1,0,1,0,1,0,1,1,1,0,1},
      {1,0,0,0,1,0,1,0,0,0,1,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,1,0,1,0,1,0,0,0,0,0,1,0,0,0,1,0,0,0,1,0,0,0,1,0,1},
      {1,0,1,1,1,0,1,0,1,1,1,0,1,1,1,0,1,1,1,1,1,1,1,0,1,1,1,0,1,0,1,1,1,1,1,0,1,1,1,1,1,1,1,1,1,1,1,0,1,0,1},
      {1,0,1,0,0,0,1,0,0,0,1,0,0,0,1,0,0,0,1,0,0,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,1,0,1},
      {1,0,1,1,1,1,1,1,1,0,1,1,1,0,1,1,1,0,1,0,1,0,1,0,1,0,1,1,1,0,1,0,1,0,1,1,1,1,1,0,1,0,1,1,1,0,1,0,1,0,1},
      {1,0,0,0,0,0,0,0,1,0,1,0,1,0,1,0,0,0,1,0,1,0,1,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,1,0,1,0,1,0,0,0,1,0,1,0,1},
      {1,0,1,1,1,1,1,0,1,0,1,0,1,0,1,0,1,1,1,0,1,0,1,0,1,0,1,1,1,1,1,0,1,0,1,0,1,1,1,0,1,0,1,0,1,1,1,1,1,0,1},
      {1,0,0,0,0,0,1,0,0,0,0,0,0,0,1,0,1,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,1,0,1,0,0,0,1,0,1,0,0,0,0,0,1,0,1},
      {1,1,1,1,1,0,1,1,1,1,1,1,1,0,1,0,1,0,1,1,1,1,1,1,1,1,1,0,1,1,1,1,1,0,1,1,1,0,1,1,1,0,1,1,1,1,1,0,1,0,1},
      {1,0,0,0,1,0,1,0,0,0,1,0,0,0,1,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,1,0,0,0,1,0,1,0,1},
      {1,0,1,1,1,0,1,1,1,0,1,0,1,1,1,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,1,1,1,0,1,0,1,0,1,0,1},
      {1,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,1,0,0,0,1},
      {1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,1}};
    boolean[][] posCheck = new boolean[maze.length][maze[0].length];
    int r = 0;
    int c = 0;
    for(int row = 0; row < maze.length; row++){
      for(int col = 0; col < maze[row].length; col++){
        if(maze[row][col]==0){
          r = row;
          c = col;
        }
      }
    }
    maze[r][c] = 3;
    mazeSolver(1, 0, maze, posCheck);
  }

  public static boolean mazeSolver(int r, int c, int[][]maze, boolean[][] posCheck){
    posCheck[r][c] = true;
    maze[r][c] = 2;

    if(maze[r][c] == 3){
      print(maze);
      return true;
    }

    if((c+1 < maze.length) && maze[r][c+1]==0 && !posCheck[r][c+1] && (mazeSolver(r, c + 1, maze, posCheck))){
      maze[r][c] = 2;
      return true;
    }

    if((r-1 >= 0) && maze[r-1][c]==0 && !posCheck[r-1][c] && (mazeSolver(r - 1, c, maze, posCheck))){
      maze[r][c] = 2;
      return true;
    }

    if((c-1 >= 0) && maze[r][c-1]==0 && !posCheck[r][c-1] && (mazeSolver(r, c - 1, maze, posCheck))){
      maze[r][c] = 2;
      return true;
    }

    if((r+1 < maze.length) && maze[r+1][c]==0 && !posCheck[r+1][c] && (mazeSolver(r + 1, c, maze, posCheck))){
      maze[r][c] = 2;
      return true;
    }

    print(maze);
    return false;
  }

  public static void print(int[][] maze){
    for(int row = 0; row<maze.length; row++){
      for(int col = 0; col<maze[row].length; col++)
        System.out.print(maze[row][col]);
      System.out.println();
    }
  }
}

2 个答案:

答案 0 :(得分:0)

让我们说你有: 初始状态

SW00WW
00W0WW
W000WW
W0WWWW
00000E
w - 墙 0 - 未走的路径 S - 起点 E - 终点 (X - 走点)

说明:

我们做什么?当我们到达叶子时,我们迭代0并用X标记它们。如果你从一点开始其他0,你只有在你确定不需要再次返回时才将其标记为X.

www00W     (see 'P' as an '0') we must go from 'S' to 'E' when we reach 'P' we have 2 moves from that point 
S00Pww      when iterating. As conclusion you let it be stil '0', and when we meet a first node 
www00E      that wont need another visit mark it as X.

示例:

SW00WW
x0W0WW
W000WW
W0WWWW
00000E

SW00WW
xxW0WW
W000WW
W0WWWW
00000E

SW00WW
xxW0WW
W000WW
W0WWWW
00000E

SW00WW
x0W0WW
W000WW
W0WWWW
00000E

SW00WW
xxW0WW
W0x0WW
W0WWWW
00000E

SW00WW
xxW0WW
W0xxWW
W0WWWW
00000E

SWxxWW   (made 3 steps on a first ipotetical choise)
xxWxWW
W0xxWW
W0WWWW
00000E

    SWxxWW   (made 4 steps on last row to the end)
    xxWxWW
    WxxxWW
    WxWWWW
    0xxxxE

希望它有所帮助,对不起很长的例子,但我试图说清楚。

PS:替代深度优先搜索

答案 1 :(得分:0)

如果将所有位置有效性测试放在一个位置,递归函数将更具“可读性”:

public static boolean mazeSolver(int r, int c, int[][]maze){
  if( ! isPositionValid(r, c, maze))
    return false;       // tried to flow outside the maze

  if(maze[r][c] == 3){  // is it a destination point?
    print(maze);        // solved
    return true;
  }

  if( maze[r][c] != 0)  // a wall, a path or already checked?
    return false;

  maze[r][c] = 2;       // mark position as a part of the path

  if( mazeSolver(r, c + 1, maze)))  // try to extend the path and
    return true;                    // return if solution found
  if( mazeSolver(r, c - 1, maze)))
    return true;
  if( mazeSolver(r + 1, c, maze)))
    return true;
  if( mazeSolver(r - 1, c, maze)))
    return true;

  maze[r][c] = 4;     // dead-end - mark the position 'checked'
  return false;
}


public static boolean isPositionValid(int r, int c, int[][]maze){
  return r >= 0 && c >= 0 && r < maze.size && c < maze[r].size;
}