Question

我对如何使用队列解决迷宫非常困惑。我已经提供了一些javadoc和一些我教授给我们的伪代码。尽可能帮助。我已经查看了其他主题，但我无法理解，希望有人可以帮我解决我的解决方法。感谢

public class QueueMazeSolver implements MazeSolver {

private MazeGUI gui;

public static class Cell {

    private int r;
    private int c;

    public Cell(int row, int col){

        r = row;
        c = col;

    }

}


public QueueMazeSolver(){

    gui = new MazeGUI( this );

}

/**
 * This method is called when the start button is
 * clicked in the MazeGUI.  This method should solve the maze.
 * This method may call MazeGUI.drawMaze(...) whenever the
 * GUI display should be updated (after each step of the solution).
 * 
 * The maze is provided as the first parameter.  It is a 2D array containing
 * characters that represent the spaces in the maze.  The following 
 * characters will be found in the array:
 *    '#' - This represents a wall.
 *    ' ' - This represents an open space (corridor)
 *    
 * When calling MazeGUI.drawMaze(...) to update the display, the GUI
 * will recognize the '#' and ' ' characters as well as the following:
 *    '@' - Means the cell is a space that has been explored
 *    '%' - Means that the cell is part of the best path to the goal.
 * 
 * @param maze the maze (see above).
 * @param startR the row of the start cell.
 * @param startC the column of the start cell.
 * @param endR the row of the end (goal) cell.
 * @param endC the column of the end (goal) cell.
 */
@Override
public void solve(char[][] maze, int startR, int startC, int endR, int endC) {

    maze[startR][startC] = '@';

    ArrayQueue<Cell> agenda = new ArrayQueue<Cell>();

    Cell temp = new Cell(startR, startC);

    agenda.offer(temp);

            // while agenda is not empty and red not found
    while(!agenda.isEmpty() && maze[endR][endC] != '@' ){

        Cell current = agenda.poll(); //remove front square from queue

        /*
        if front square is red
            found it
        else 
            mark amaze all unexplored neighbors of front
            square and add them to the square
        */

        if(current == new Cell(endR, endC) ){
            break;
        }
        else{

            =
            }
        }

        /** Notes
        maze[r][c] = '@' //marking cell seen

        up = r-1, c
        down = r+1, c
        left = r, c-1
        right = r, c+1
        */


    }

    if (!agenda.isEmpty())
            gui.setStatusText("Maze is solvable");
    else
        gui.setStatusText("Maze is unsolvable");



    gui.drawMaze(maze);
    try {Thread.sleep(150);}
    catch (InterruptedException e){
        System.err.println("Thread interrupted");
    }       
 }

public static void main(String[] args){

    QueueMazeSolver solver = new QueueMazeSolver();

   }

     }

Answer 1

似乎你正试图获得可能在迷宫中移动并到达red单元格的路径，这些单元格有墙（无法交叉）或开放空间。

此代码基本上是应用breadth first search 我们从queue移除了一个单元格，如果未访问周围的单元格[at distance 1 unit]，请将其添加到queue并访问它们。
伪代码（来自维基百科）：

1  procedure BFS(G,v) is
2      create a queue Q
3      create a vector set V
4      enqueue v onto Q
5      add v to V
6      while Q is not empty loop
7         t ← Q.dequeue()
8         if t is what we are looking for then
9            return t
10        end if
11        for all edges e in G.adjacentEdges(t) loop
12           u ← G.adjacentVertex(t,e)
13           if u is not in V then
14               add u to V
15               enqueue u onto Q
16           end if
17        end loop
18     end loop
19     return none
20 end BFS

假设您位于小区(i,j)，因此t = (i,j)且adjacentEdges（t）为(i+1,j)，(i,j+1)，(i-1,j)。 (i,j-1) 如果以前没有访问(i+1,j)，请将其添加到队列中（因此，下次从队列中弹出时，您将获得它）否则如果访问它（即在V中）则我们完成了它。对其他三个细胞重复相同的操作

通过这种方式，您可以执行O(m*n)次操作，并且只需访问每个单元格一次。

如何用队列解决迷宫？

1 个答案: