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我在哪里放置边界检查以便程序生成整个迷宫?代码应该打印一个网格,其中包含通过打破单元格之间的墙壁绘制的迷宫。然而,令我沮丧的是,网格在到达索引 0 或 24 时停止。我需要程序在它停止之前访问每个单元格(如果它进入边界,它会向后移动)。

这是我得到的上一个错误:

Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: -1
        at Grid.genRand(Grid.java:73)
        at Grid.main(Grid.java:35)

这是源代码:

import java.awt.*;
import java.awt.Color;
import java.awt.Component;
import java.awt.Graphics;
import java.util.ArrayList;

public class Grid extends Canvas {

    Cell[][] maze;
    int size;
    int pathSize;
    double width, height;
    ArrayList<int[]> coordinates = new ArrayList<int[]>();

    public Grid(int size, int h, int w) {
        this.size = size;
        maze = new Cell[size][size];
        for(int i = 0; i<size; i++){
            for(int a =0; a<size; a++){
            maze[i][a] = new Cell();
            }
        }
        setPreferredSize(new Dimension(h, w));
    }

    public static void main(String[] args) {
        Frame y = new Frame();
        y.setLayout(new BorderLayout());
        Panel r = new Panel();
        r.setLayout(new BorderLayout());
        Grid f = new Grid(25, 400, 400);
        r.add(f, BorderLayout.CENTER);
        y.add(r, BorderLayout.CENTER);
        f.genRand();
        f.repaint();
        y.pack();
        y.setPreferredSize(new Dimension(450, 450));
        y.setVisible(true);
    }

    public void push(int[] xy){
        coordinates.add(xy);
        int i = coordinates.size();
        coordinates.ensureCapacity(i++);
    }

    public int[] pop(){
        int[] x = coordinates.get((coordinates.size())-1);
        coordinates.remove((coordinates.size())-1);
        return x;
    }

    public int[] top(){
        return coordinates.get((coordinates.size())-1);
    }

    public void genRand(){
        // create a CellStack (LIFO) to hold a list of cell locations [x]
        // set TotalCells = number of cells in grid  
        int TotalCells = size*size;
        // choose a cell at random and call it CurrentCell 
        int m = randomInt(size);
        int n = randomInt(size);
        while(m<1){
            m = randomInt(size);
        }
        while(n<1){
            n = randomInt(size);
            }
        Cell curCel = maze[m][n];
        // set VisitedCells = 1  
        int visCel = 1;
        int o = 0;
        int p = 0;
        int h;
        int d;
        int[] q;
        // while VisitedCells < TotalCells 
        while( visCel < TotalCells){
            d = 0;
            // find all neighbors of CurrentCell with all walls intact
            if(m!=0&&n!=0){
                if(m<size&&n<size){
                    if(maze[m-1][n].countWalls() == 4)
                        {d++;}
                    if(maze[m+1][n].countWalls() == 4)
                        {d++;}
                    if(maze[m][n-1].countWalls() == 4)
                        {d++;}
                    if(maze[m][n+1].countWalls() == 4)
                        {d++;}
                }
            }
            // if one or more found 
            if(d!=0){
                Point[] ls = new Point[4];
                ls[0] = new Point(m-1,n);
                ls[1] = new Point(m+1,n);
                ls[2] = new Point(m,n-1);
                ls[3] = new Point(m,n+1);
                // knock down the wall between it and CurrentCell
                h = randomInt(3);
                switch(h){
                    case 0: o = (int)(ls[0].getX());
                            p = (int)(ls[0].getY());
                            curCel.destroyWall(2);
                            maze[o][p].destroyWall(1);
                        break;
                    case 1: o = (int)(ls[1].getX());
                            p = (int)(ls[1].getY());
                            curCel.destroyWall(1);
                            maze[o][p].destroyWall(2);
                        break;
                    case 2: o = (int)(ls[2].getX());
                            p = (int)(ls[2].getY());
                            curCel.destroyWall(3);
                            maze[o][p].destroyWall(0);
                        break;
                    case 3: o = (int)(ls[3].getX());
                            p = (int)(ls[3].getY());
                            curCel.destroyWall(0);
                            maze[o][p].destroyWall(3);
                        break;
                }   
                // push CurrentCell location on the CellStack 
                push(new int[] {m,n});
                // make the new cell CurrentCell
                m = o;
                n = p;
                curCel = maze[m][n];
                // add 1 to VisitedCells
                visCel++;
            }
            // else 
            else{
                // pop the most recent cell entry off the CellStack 
                q = pop();
                m = q[0];
                n = q[1];
                curCel = maze[m][n]; 
                // make it CurrentCell
                // endIf
            }
        // endWhile  
        }   
    }

    public int randomInt(int s) { return (int)(s* Math.random());}

    public void paint(Graphics g) {
        int k, j;
        width = getSize().width;
        height = getSize().height;
        double htOfRow = height / (size);
        double wdOfRow = width / (size);
//checks verticals - destroys east border of cell
        for (k = 0; k < size; k++) {
            for (j = 0; j < size; j++) {
                if(maze[k][j].checkWall(2)){
                g.drawLine((int) (k * wdOfRow), (int) (j * htOfRow), (int) (k * wdOfRow), (int) ((j+1) * htOfRow));
            }}
        }
//checks horizontal - destroys north border of cell
        for (k = 0; k < size; k++) {
            for (j = 0; j < size; j++) {
                if(maze[k][j].checkWall(3)){
                g.drawLine((int) (k * wdOfRow), (int) (j * htOfRow), (int) ((k+1) * wdOfRow), (int) (j * htOfRow));
            }}
        }
    }
}

class Cell {

    private final static int NORTH = 0;
    private final static int EAST = 1;
    private final static int WEST = 2;
    private final static int SOUTH = 3;
    private final static int NO = 4;
    private final static int START = 1;
    private final static int END = 2;
    boolean[] wall = new boolean[4];
    boolean[] border = new boolean[4];
    boolean[] backtrack = new boolean[4];
    boolean[] solution = new boolean[4];
    private boolean isVisited = false;
    private int Key = 0;

    public Cell(){
    for(int i=0;i<4;i++){wall[i] = true;}
    }
    public int countWalls(){
    int i, k =0; 
    for(i=0; i<4; i++) {
    if (wall[i] == true)
    {k++;}
    }
    return k;}
    public boolean checkWall(int x){
    switch(x){
        case 0: return wall[0];
        case 1: return wall[1];
        case 2: return wall[2];
        case 3: return wall[3];
    }
    return true;
    }
    public void destroyWall(int x){
    switch(x){
        case 0: wall[0] = false; break;
        case 1: wall[1] = false; break;
        case 2: wall[2] = false; break;
        case 3: wall[3] = false; break;
        }
    }
    public void setStart(int i){Key = i;}   
    public int getKey(){return Key;}
    public boolean checkVisit(){return isVisited;}
    public void visitCell(){isVisited = true;}
}
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1 回答 1

0

我很快就忽略了你的代码:

  1. 当坐标为空时调用 top() 或 pop() 时可能会出现异常 --> coordinate.get(coordinates.size()-1) 如果大小为 0,则尝试寻址索引 -1 -> BAM!

  2. 有些方法看起来确实很复杂。您可以将该功能拆分为单独的方法/功能

  3. 您不要混合 AWT 和 SWING 组件,因为它们的行为不同

  4. 永远不要在事件调度程序线程 (EDT) 之外执行 GUI 操作(显示、隐藏、更改内容、单击、JFrame、JPanel 等中的任何内容)-> 这可能会导致死锁

于 2012-04-09T14:55:46.277 回答