使Java递归迷宫求解器更高效
[英]Make Java recursive maze solver more efficient
问题描述
更新
我可以通过将线程大小增加到几GB来使我的算法正常工作，并且在一两秒钟之内就能解决1803x1803迷宫问题。
---------------
我昨天开始用Java教自己的递归。 我创建了一种算法，可以为迷宫拍照并解决。 但是，当做大于200x200 px的迷宫时，我会得到堆栈溢出的答案，因为我认为此算法的堆栈变得太长。 如何改善此算法，以便可以输入最大1000x1000的图像？
另外，您能告诉我我目前使用哪种算法吗？ 我相信这是DFS，但我不确定。
请解释为什么您的解决方案更有效，以及使用的想法。
这是解决问题的主要方法
public class BlackWhiteSolver {
static int[][] solutionSet = new int[203][203];
static int width, height;
static String originalImage;
static int correctX, correctY;
public static void convert() {
try {
BufferedImage original = ImageIO.read(new File(originalImage));
int red;
int threshold = 2;
width = original.getWidth();
height = original.getHeight();
for(int i=0; i<original.getWidth(); i++) {
for(int j=0; j<original.getHeight(); j++) {
red = new Color(original.getRGB(i, j)).getRed();
// 1 = white, 0 = black, 9 = tried, 5 = solved
if(red > threshold) { solutionSet[i][j] = 1; }
else { solutionSet[i][j] = 0; }
}
}
} catch (IOException e) {e.printStackTrace();}
}
public BlackWhiteSolver(int solvedX, int solvedY, String pic) {
correctX = solvedX;
correctY = solvedY;
originalImage = pic;
}
public boolean solve (int row, int column) {
boolean completed = false;
if (validPoint(row, column)) {
solutionSet[row][column] = 9;
if (row == correctX && column == correctY) {
completed = true;
} else {
completed = solve (row+1, column);
if (!completed) {
completed = solve (row, column+1);
}
if (!completed) {
completed = solve (row-1, column);
}
if (!completed) {
completed = solve (row, column-1);
}
}
if (completed) {
solutionSet[row][column] = 5;
}
}
return completed;
}
private boolean validPoint (int row, int column) {
boolean isValid = false;
if (row < height-1 && column < width-1 && row >= 1 && column >= 1 ) {
if (solutionSet[row][column] == 1) {
isValid = true;
}
}
return isValid;
}
public static void solvedFile() {
BufferedImage binarized = new BufferedImage(width, height,BufferedImage.TYPE_3BYTE_BGR);
int newPixel = 0;
int rgb = new Color(255, 0, 0).getRGB();
for(int i=0; i<width; i++){
for(int j=0; j<height; j++)
{
if (solutionSet[i][j] == 0) {
newPixel = 0;
newPixel = colorToRGB(1, newPixel, newPixel, newPixel);
} else if (solutionSet[i][j] == 1 || solutionSet[i][j] == 9) {
newPixel = 255;
newPixel = colorToRGB(1, newPixel, newPixel, newPixel);
} else if (solutionSet[i][j] == 5) {
newPixel = 16711680;
}
binarized.setRGB(i, j, newPixel);
}
}
try { ImageIO.write(binarized, "gif",new File("maze-complete") );} catch (IOException e) {e.printStackTrace();}
}
private static int colorToRGB(int alpha, int red, int green, int blue) {
int newPixel = 0;
newPixel += alpha;
newPixel = newPixel << 8;
newPixel += red; newPixel = newPixel << 8;
newPixel += green; newPixel = newPixel << 8;
newPixel += blue;
return newPixel;
}
}
这是跑迷宫的课
public class BlackWhiteInterface
{
public static void main (String[] args) {
BlackWhiteSolver puzzle = new BlackWhiteSolver(60, 202, "maze-4.gif");
System.out.println();
puzzle.convert();
if (puzzle.solve(0,34)) {
System.out.println("completed");
puzzle.solvedFile();
} else {
System.out.println("not possible");
}
}
}
生成带有起点和终点的正确迷宫
public class MazeBuilder {
static String start = "left";
static String end = "down";
public static void main(String[] args)
{
try
{
BufferedImage original = ImageIO.read(new File("mazeInput1.gif"));
BufferedImage binarized = new BufferedImage(original.getWidth(), original.getHeight(),BufferedImage.TYPE_BYTE_BINARY);
int red;
int redRightPixel;
int redUpPixel;
int newPixel;
int threshold = 2;
for(int i=0; i<original.getWidth(); i++)
{
for(int j=0; j<original.getHeight(); j++)
{
red = new Color(original.getRGB(i, j)).getRed();
int alpha = new Color(original.getRGB(i, j)).getAlpha();
if(red > threshold) { newPixel = 255; }
else { newPixel = 0; }
if (i == 0 || j == 0 || i == original.getWidth()-1 || j == original.getHeight() - 1){
newPixel = 0;
if (end == "left") {
} else if (end == "right") {
} else if (end == "up") {
} else if (end == "down") {
}
/*if (i == 1 || j == 1 || i == original.getWidth()-2 || j == original.getHeight() - 2 && red > 2) {
System.out.println("Start Point: (" + i + ", " + j + ")");
}
if (i == 0 && j > 0 && j < original.getHeight()-1) {
redRightPixel = new Color(original.getRGB(i+1, j)).getRed();
if (i == 0 && redRightPixel > 2) {
System.out.println("Start Point: (" + i + ", " + j + ")");
newPixel = 255;
}
}*/
/*if (j == original.getHeight()-1 && i > 0 && i < original.getWidth()-1) {
redUpPixel = new Color(original.getRGB(i, j-1)).getRed();
if (redUpPixel > 2) {
System.out.println("End Point: (" + i + ", " + j + ")");
newPixel = 255;
}
}*/
}
if (start == "left") {
if (i == 1 && j != 0 && j != original.getHeight()-1 && red > 2) {
System.out.println("Start Point: (" + i + ", " + j + ")");
}
} else if (start == "right") {
if (i == original.getHeight()-2 && j != 0 && j != original.getHeight()-1 && red > threshold) {
System.out.println("Start Point: (" + i + ", " + j + ")");
}
} else if (start == "up") {
if (j == 1 && i != 0 && i != original.getWidth()-1 && red > threshold) {
System.out.println("Start Point: (" + i + ", " + j + ")");
}
} else if (start == "down") {
if (j == original.getHeight()-2 && i != 0 && i != original.getWidth()-1 && red > threshold) {
System.out.println("Start Point: (" + i + ", " + j + ")");
}
}
if (end == "left") {
if (i == 1 && j != 0 && j != original.getHeight()-1 && red > 2) {
System.out.println("End Point: (" + i + ", " + j + ")");
}
} else if (end == "right") {
if (i == original.getHeight()-2 && j != 0 && j != original.getHeight()-1 && red > threshold) {
System.out.println("End Point: (" + i + ", " + j + ")");
}
} else if (end == "up") {
if (j == 1 && i != 0 && i != original.getWidth()-1 && red > threshold) {
System.out.println("End Point: (" + i + ", " + j + ")");
}
} else if (end == "down") {
if (j == original.getHeight()-2 && i != 0 && i != original.getWidth()-1 && red > threshold) {
System.out.println("End Point: (" + i + ", " + j + ")");
}
}
newPixel = colorToRGB(alpha, newPixel, newPixel, newPixel);
binarized.setRGB(i, j, newPixel);
}
}
ImageIO.write(binarized, "gif",new File("maze-4") );
}
catch (IOException e)
{
e.printStackTrace();
}
}
private static int colorToRGB(int alpha, int red, int green, int blue) {
int newPixel = 0;
newPixel += alpha;
newPixel = newPixel << 8;
newPixel += red; newPixel = newPixel << 8;
newPixel += green; newPixel = newPixel << 8;
newPixel += blue;
return newPixel;
}
}
203 x 203迷宫的示例输出
1 个解决方案
解决方案1
0 2018-09-27 00:41:07
一种稍微有效的简单方法是不使用递归将到目前为止已遵循的路径存储在堆栈中。 取而代之的是将您到目前为止遵循的路径存储在java.util.BitSet （在其中将每个路径像素存储在BitSet y*width + x元素中），或者您可以简单地使用图片的红色区域上色以存储路径。
这样可以避免堆栈溢出。
基本算法是从起点开始，并沿四个主要方向之一进行，除非您已经访问过该方向（尝试过并找到一个死胡同，或者从该方向来到这里）。 当您朝某个方向行驶时，您在那做同样的事情。 这是一个简单的非递归循环。
当您走到尽头时，可以通过检查所有四个方向（从何处看）来弄清楚最初是如何到达那里的。 您将红色从站立的位置上移开，然后沿原来的方向返回。 如果在任何方向上都没有红色路径，那么您将再次处于起点，并且您已尝试了一切，因此无法解决迷宫问题。
当您回溯时，您可以在路径上的较旧的正方形上尝试尚未尝试的下一个方向，直到所有方向都死胡同为止。
如果您到达终点，那就完成了。
这是一些通常无法处理循环的伪代码（路径在“圆圈”中），效率极低（例如，它应使用BitSet而不是boolean[][] ），并且可能存在一些错误，但它给出了大致的想法：
public class MazeSolver {
private static enum Direction { UP, RIGHT, DOWN, LEFT }
// Return array's element is true if that's part of the path
public static boolean[][] solve(final boolean[][] mazeWallHere,
int x, int y,
final int endX, final int endY) {
final int width = mazeWallHere.length;
final int height = mazeWallHere[0].length;
final boolean[][] path = new boolean[width][height];
Direction nextDirection = Direction.UP;
boolean backtrack = false;
while (true) {
// If this spot is a dead end in all new directions, head back
if (backtrack) {
backtrack = false;
// Unmark where we are
path[x][y] = false;
// Find where we came from and what direction we took to get here
// Then switch to the next direction
// If all directions have been tried, backtrack again
// If we can't backtrack, return null because there's no solution
// If we went up to get here, go back down and try going right.
if (y != 0 && path[x][y - 1]) {
y--;
nextDirection = Direction.RIGHT;
continue;
}
// If we went right to get here, go back left and try going down.
else if (x != 0 && path[x - 1][y]) {
x--;
nextDirection = Direction.DOWN;
continue;
}
// If we went down to get here, go back up and try going left.
else if (y < height && path[x][y + 1]) {
y++;
nextDirection = Direction.LEFT;
continue;
}
// If we went left to get here, go back right and backtrack again.
else if (x < width && path[x + 1][y]) {
x++;
backtrack = true;
continue;
}
// If we didn't come from anywhere, we're at the starting point
// All possible paths are dead ends
else return null;
}
// Mark where we are
path[x][y] = true;
// If we've solved it, return the solution
if (x == endX && y == endY) return path;
// Move unless we:
// * hit the edge of the maze
// * it's the direction we originally got here from
// * hit a wall
// If we can't go a certain direction, try the next direction
// If we're out of directions to try, backtrack
switch (nextDirection) {
case UP: if (y == height
|| path[x][y + 1]
|| mazeWallHere[x][y + 1]) {
nextDirection = Direction.RIGHT;
continue;
}
else y++;
break;
case RIGHT: if (x == width
|| path[x + 1][y]
|| mazeWallHere[x + 1][y]) {
nextDirection = Direction.DOWN;
continue;
}
else x++;
break;
case DOWN: if (y == 0
|| path[x][y - 1]
|| mazeWallHere[x][y - 1]) {
nextDirection = Direction.LEFT;
continue;
}
else y--;
break;
case LEFT: if (x == 0
|| path[x - 1][y]
|| mazeWallHere[x - 1][y]) {
backtrack = true;
continue;
}
else x--;
break;
}
}
}
}
如果要正确处理循环，请将path设为int[][]并存储移动号而不是true，以便您知道哪个路径更旧。
Java递归迷宫求解器问题
声明:本站的技术帖子网页，遵循CC BY-SA 4.0协议，如果您需要转载，请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.