使Java递归迷宫求解器更高效
更新
我能够通过将线程大小增加到几GB来使算法工作，并且能够在一两秒钟内解决1803x1803迷宫。
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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？
此外，您能告诉我我目前正在使用哪种算法吗？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其使用的想法。
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;
}
}
这是运行迷宫的课程
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");
}
}
}
使用起点和终点生成正确的迷宫
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 个答案:
答案 0 :(得分：0)
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中（您将每个路径像素存储在y*width + x的元素BitSet中），也可以只使用红色的用来存储路径的彩色图片。
这避免了堆栈溢出。
基本算法是从起点开始，并沿着四个基本方向之一进行，除非您已经访问过该方向（要么尝试并找到一个死胡同，要么从该方向来到这里） 。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，您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，并且您已经尝试了一切，因此无法解决迷宫。
回溯时，您会尝试在路径上的较旧广场尝试尚未尝试过的下一个方向，直到所有方向都死胡同为止。
如果您到达终点，就好了。
这是一些伪代码，它们通常不能处理循环（路径在“圆圈”中），效率极低（例如，应使用BitSet而不是boolean[][]），可能有一些错误，但这给出了大致的想法：
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;
}
}
}
}
如果您要正确处理循环，请将path设为int[][]并存储移动编号（而不是true），以便知道哪个路径更旧。
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