我正在开发适用于Android的游戏，其中可随机生成可探索区域。现在我只是试图让迷宫产生（有一些ASCII艺术输出，所以我可以看到它），而且我已经在这里待了大约4-5天，但我只是难住了。
我试图使用＆＃34;深度优先搜索＆＃34;算法，我能找到的所有例子都使用递归回溯。由于这是针对Android而且手机相对比较懦弱，递归会很快导致调用堆栈溢出，这就是我尝试使用堆栈进行回溯编写自己的算法的原因。
我想出了这个解决方案，使用MazeGenerator类和MazeCell类。
MazeGenerator：
package com.zarokima.mistwalkers.explore;
import java.util.Random;
import java.util.Stack;
import org.anddev.andengine.util.path.Direction;
import android.graphics.Point;
public class MazeGenerator
{
private int x, y; // the dimensions of the maze
private MazeCell[][] maze;
private Random rand = new Random();
private Stack<MazeCell> stack;
public MazeGenerator(int x, int y)
{
this.x = x;
this.y = y;
generateMaze();
}
public void setSeed(long seed)
{
rand.setSeed(seed);
}
public void setSize(int x, int y)
{
this.x = x;
this.y = y;
}
public String outputMazeText()
{
String output = new String();
for (int i = 0; i < y; i++)
{
// draw the north edge
for (int k = 0; k < x; k++)
{
output += maze[k][i].hasNeighbor(Direction.UP) ? "+ " : "+---";
}
output += "+\n";
// draw the west edge
for (int k = 0; k < x; k++)
{
output += maze[k][i].hasNeighbor(Direction.LEFT) ? " " : "| ";
}
output += "|\n";
}
// draw the bottom line
for (int k = 0; k < x; k++)
{
output += "+---";
}
output += "+\n";
return output;
}
public void generateMaze()
{
maze = new MazeCell[x][y];
for (int i = 0; i < x; i++)
{
for (int k = 0; k < y; k++)
{
maze[i][k] = new MazeCell(i, k);
}
}
MazeCell.setBounds(x, y);
stack = new Stack<MazeCell>();
stack.push(maze[0][0]);
maze[0][0].setInMaze(true);
while (!stack.isEmpty())
{
MazeCell currentCell = stack.peek();
Direction[] possibleDirections = currentCell.getUncheckedDirections();
if (possibleDirections.length == 0)
{
stack.pop();
continue;
}
int dint = rand.nextInt(possibleDirections.length);
Direction direction = possibleDirections[dint];
MazeCell nextCell = null;
Point position = currentCell.getPosition();
switch (direction)
{
case UP:
nextCell = maze[position.x][position.y - 1];
break;
case DOWN:
nextCell = maze[position.x][position.y + 1];
break;
case LEFT:
nextCell = maze[position.x - 1][position.y];
break;
case RIGHT:
nextCell = maze[position.x + 1][position.y];
break;
}
currentCell.setNeighbor(nextCell, direction);
stack.push(nextCell);
}
}
}
MazeCell：
package com.zarokima.mistwalkers.explore;
import java.util.ArrayList;
import org.anddev.andengine.util.path.Direction;
import android.graphics.Point;
public class MazeCell
{
private MazeCell[] neighbors;
private boolean[] checked;
private boolean inMaze = false;
private Point position;
private static boolean setNeighbor = true; //whether the next call of SetNeighbor() should also call for the new neighbor
private static int xMax = 10, yMax = 10; //exclusive boundary for position
private int mapIndex; //will be used when maze generation is working properly
public MazeCell(int x, int y)
{
position = new Point(x,y);
neighbors = new MazeCell[4];
checked = new boolean[4];
for(int i = 0; i < neighbors.length; i++)
{
neighbors[i] = null;
}
}
public Point getPosition()
{
return position;
}
public void setInMaze(boolean b)
{
inMaze = b;
}
public static void setBounds(int x, int y)
{
xMax = x;
yMax = y;
}
public void setNeighbor(MazeCell c, Direction d)
{
checked[d.ordinal()] = true;
switch(d)
{
case UP:
if(!c.hasNeighbor(Direction.DOWN) && !c.isInMaze());
{
if(setNeighbor)
{
setNeighbor = false;
c.setNeighbor(this, Direction.DOWN);
}
neighbors[d.ordinal()] = c;
}
break;
case DOWN:
if(!c.hasNeighbor(Direction.UP) && !c.isInMaze())
{
if(setNeighbor)
{
setNeighbor = false;
c.setNeighbor(this, Direction.UP);
}
neighbors[d.ordinal()] = c;
}
break;
case LEFT:
if(!c.hasNeighbor(Direction.RIGHT) && !c.isInMaze())
{
if(setNeighbor)
{
setNeighbor = false;
c.setNeighbor(this, Direction.RIGHT);
}
neighbors[d.ordinal()] = c;
}
break;
case RIGHT:
if(!c.hasNeighbor(Direction.LEFT) && !c.isInMaze())
{
if(setNeighbor)
{
setNeighbor = false;
c.setNeighbor(this, Direction.LEFT);
}
neighbors[d.ordinal()] = c;
}
break;
}
setNeighbor = true;
inMaze = true;
}
public void setDirectionChecked(Direction d, boolean b)
{
checked[d.ordinal()] = b;
}
public boolean hasNeighbor(Direction d)
{
return (neighbors[d.ordinal()] != null);
}
public MazeCell getNeighbor(Direction d)
{
return neighbors[d.ordinal()];
}
public boolean isInMaze()
{
return inMaze;
}
public Direction[] getUncheckedDirections()
{
ArrayList<Direction> al = new ArrayList<Direction>();
for(Direction d : Direction.values())
{
//boundary cases
switch(d)
{
case UP:
if(position.y == 0)
continue;
break;
case DOWN:
if(position.y == yMax-1)
continue;
break;
case LEFT:
if(position.x == 0)
continue;
break;
case RIGHT:
if(position.x == xMax-1)
continue;
break;
}
if(checked[d.ordinal()] == false)
al.add(d);
}
Direction[] d = new Direction[al.size()];
for(int i = 0; i < d.length; i++)
d[i] = al.get(i);
return d;
}
}
这会产生类似于this
的结果注意每个单元格如何始终连接到其上下邻居。我一直无法弄清楚这里有什么问题。
虽然MazeCell的setNeighbor功能中的检查看起来应该足够了，但我还添加了一些内容，看看会发生什么。这是第二个generateMaze（）方法：
public void generateMaze()
{
maze = new MazeCell[x][y];
for (int i = 0; i < x; i++)
{
for (int k = 0; k < y; k++)
{
maze[i][k] = new MazeCell(i, k);
}
}
MazeCell.setBounds(x, y);
stack = new Stack<MazeCell>();
stack.push(maze[0][0]);
maze[0][0].setInMaze(true);
while (!stack.isEmpty())
{
MazeCell currentCell = stack.peek();
Direction[] possibleDirections = currentCell.getUncheckedDirections();
if (possibleDirections.length == 0)
{
stack.pop();
continue;
}
int dint = rand.nextInt(possibleDirections.length);
Direction direction = possibleDirections[dint];
currentCell.setDirectionChecked(direction, true);
MazeCell nextCell = null;
Point position = currentCell.getPosition();
switch (direction)
{
case UP:
nextCell = maze[position.x][position.y - 1];
break;
case DOWN:
nextCell = maze[position.x][position.y + 1];
break;
case LEFT:
nextCell = maze[position.x - 1][position.y];
break;
case RIGHT:
nextCell = maze[position.x + 1][position.y];
break;
}
if (!nextCell.isInMaze())
{
currentCell.setNeighbor(nextCell, direction);
stack.push(nextCell);
}
}
它产生的结果如this
注意细分如何分解。
我已经玩了很多很多，而不仅仅是这里提到的，但没有任何显示任何真正的改进 - 大多数最终只是看起来像第二张图片。有什么帮助吗？
答案 0 :(得分：1)
我建议创建一个名为Direction oppositeOf(Direction d)的函数（具有明显的逻辑）。此函数允许您在setNeighbor中完全删除switch语句（如果已添加）。
在这里，我重写setNeighbor以获得与上面完全相同的逻辑，只需使用此函数：
public void setNeighbor(MazeCell c, Direction d)
{
checked[d.ordinal()] = true;
if (!c.isInMaze() && !c.hasNeighbor(oppositeOf(d)))
{
if (setNeighbor)
{
setNeighbor = false;
c.setNeighbor(this, oppositeOf(d));
}
neighbors[d.ordinal()] = c;
{
setNeighbor = true;
inMaze = true;
}
...实际上暴露了setNeighbor boolean 总是等于true（不管它是否设置为false，它总是设置为true），我愿意打赌你不希望它这样做。
这可能不是您最大的问题，可能还有其他逻辑错误。
答案 1 :(得分：1)
我认为您找到的递归算法很好。您只需要使用堆栈或队列而不是递归调用（模拟调用堆栈）将它们转换为迭代的。您可以找到breadth first迭代here的一个很好的示例。希望这会有所帮助，您可以根据自己的问题进行调整。