联合查找2d数组(Java)
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我正在研究一个可以识别图像形状的程序,我已经能够将图像更改为二进制值0 / 1,0 +黑色,1 =白色,然后将这些值存储到[的二维数组中。 height] [width]索引。
我正在尝试搜索数组并根据它们是否触摸另一个白色像素来合并白色像素。如果它们是,那么计数器将计算每个单独的白色像素簇,以便程序可以打印出有多少总簇。
我的代码到目前为止:
import javafx.stage.FileChooser;
import javax.imageio.ImageIO;
import javax.swing.*;
import javax.swing.text.html.ImageView;
import java.awt.*;
import java.awt.image.BufferedImage;
import java.io.File;
import java.io.IOException;
import java.util.Arrays;
import java.util.BitSet;
public class Image {
ImageView myImageView;
public static void main(String args[])
{
new Image();
}
public Image()
{
EventQueue.invokeLater(new Runnable() {
@Override
public void run() {
try {
UIManager.setLookAndFeel(UIManager.getSystemLookAndFeelClassName());
} catch (Exception ex) {
}
JFrame frame = new JFrame("Image");
frame.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE); //closes application properly
frame.add(new ImagePane());
frame.pack();
frame.setLocationRelativeTo(null);
frame.setVisible(true);
}
});
}
public static class ImagePane extends JPanel {
ImageView myImageView;
private BufferedImage image;
private BufferedImage bwImage;
int width;
int height;
int wh = width * height;
int g;
int h = 1;
BitSet bits = new BitSet(wh);
boolean[][] visited;
int[][] picture;
int[][] arr;
public ImagePane() {
try {
FileChooser fileChooser = new FileChooser();
image = ImageIO.read(new File("C:/Users/Connor/Desktop/image.png"));
this.image = image;
bwImage = new BufferedImage(image.getWidth(), image.getHeight(), BufferedImage.TYPE_BYTE_BINARY);
this.bwImage = bwImage;
Graphics2D g2d = bwImage.createGraphics();
g2d.drawImage(image, 0, 0, this);
g2d.dispose();
} catch (IOException ex) {
ex.printStackTrace();
}
// PRINTS POS OF WHITE PIXELS
int width = bwImage.getWidth();
this.width = width;
int height = bwImage.getHeight();
this.height = height;
int[][] arr = new int[height][width];
DisjointSet ds = new DisjointSet();
for (int y = 0; y < height; ++y) {
for (int x = 0; x < width; ++x) {
if (bwImage.getRGB(x, y) == 0xffffffff) {
//bits.set((y * width + x));
System.out.print("1");
arr[y][x] = y+1;
int i = 1;
ds.makeSet(y);
} else {
arr[y][x] = 0;
System.out.print("0");
ds.makeSet(0);
}
}
System.out.println("");
/*
for(int d = 0; d < height; ++d) {
ds.union(g, h);
g++;
h++;
}
*/
}
System.out.println("");
System.out.println(Arrays.deepToString(arr)); //print 2d array
System.out.println("");
} // END OF IMAGEPANE
public int getBitIndex(int position)
{
bits.get(0, width*height);
return position;
}
public Dimension getPreferredSize() {
Dimension size = super.getPreferredSize();
if (image != null) {
size = new Dimension(image.getWidth() * 2, image.getHeight());
}
return size;
}
@Override
protected void paintComponent(Graphics g) {
super.paintComponent(g);
if (image != null) {
int x = (getWidth() - (image.getWidth() * 2)) / 2;
int y = (getHeight() - (image.getHeight()));
g.drawImage(image, x, y, this);
x += image.getWidth();
g.drawImage(bwImage, x, y, this);
}
}
} // end of JPanel//
} //end of class
和不相交的集合/联合发现:
import java.util.HashMap;
import java.util.Map;
public class DisjointSet {
private Map<Integer, Node> map = new HashMap<>();
class Node {
int data;
Node parent;
int rank;
}
//Create a set with one element
public void makeSet(int data) {
Node node = new Node();
node.data = data;
node.parent = node;
node.rank = 0;
map.put(data, node);
}
//Combine sets together - Union by rank
public boolean union(int data1, int data2) {
Node node1 = map.get(data1);
Node node2 = map.get(data2);
Node parent1 = findSet(node1);
Node parent2 = findSet(node2);
if(parent1.data == parent2.data) { //if part of the same set do nothing
return false;
}
if(parent1.rank >= parent2.rank) { //highest rank becomes parent
parent1.rank = (parent1.rank == parent2.rank) ? parent1.rank +1 : parent1.rank;
parent2.parent = parent1;
} else {
parent1.parent = parent2;
} return true;
}
//find the rep of the set
public int findSet(int data) {
return findSet(map.get(data)).data;
}
//find rep recursively and path compression
private Node findSet(Node node) {
Node parent = node.parent;
if(parent == node) {
return parent;
}
node.parent = findSet(node.parent);
return node.parent;
}
}
我之前没有做过这样的事情所以我不完全确定要做什么,我想我需要检查所有方向(NESW)以查看像素是否匹配,但我没有任何运气到目前为止工作,所以我想知道有更简单的方法吗?
答案
为了演示如何在2D阵列上执行Union-Find(确定连接组件),编写了以下测试用例(见下文)
每个元素的节点Id由方法getId()唯一定义,并基于行和列位置(0,1,2 ......)。该算法从左上角(0,0)开始逐步通过2D阵列。如果组件是相同的(零或一个),则为它的东邻居及其南邻居建立联盟。边界条件包括在迭代最后一行或最后一列时不检查其中一个邻居。最后,所有邻居都将被检查,并且你留下了零和一的树。您的UF实施必须检查创建的白色群集。如果您有兴趣,我可以发布我的UF实施。
您的问题更多地涉及如何迭代2D矩阵,如下所示:
public class UnionFindArray {
@Test
public void test() {
int a[][] = {{ 1, 0, 0, 1 }, {1, 0, 0, 1}, {0, 1, 1, 0}, {0, 1, 1, 0}};
int nRows = a.length;
int nCols = a[0].length;
UnionFind uf = new QuickUnionFind(nRows*nCols); // This is my implementation - you need to substitute yours
// Examine all neighbours
for (int row = 0; row < nRows; row++) {
for (int col = 0; col < nCols; col++) {
if (row < nRows-1 && a[row][col]==a[row+1][col])
uf.union(getId(row,col,nCols), getId(row+1,col,nCols));
if (col < nCols-1 && a[row][col]==a[row][col+1])
uf.union(getId(row,col,nCols), getId(row,col+1,nCols));
}
}
assertEquals(6, uf.getNumberOfTrees()); // True - there are 3 trees of zeros, and 3 trees of ones
}
private int getId(int row, int col, int nCols) {
return row*nCols + col;
}
}
接口:
public interface UnionFind {
public void union(int p, int q);
public boolean connected(int p, int q);
public int getNumberOfTrees();
}
执行:
/**
* Uses weighting and path compression to improve efficiency
* The nodes will form trees; each connected node will point to
* a parent. In this way, union is of order O(1).
* To keep the find as close to O(1) as possible, the tree
* must stay flat.
* To keep it flat:
* 1) Use weighting: merge the SMALLER tree into the larger
* 2) Use path compression; during the find(root), use the
* fact that you are traversing the tree to help flatten it.
* @author Ian McMaster
*
*/
public class QuickUnionFind implements UnionFind {
private final int N; // Number of nodes in the forest
private final int id[]; // The parent of node n (a node in a tree)
private final int sz[]; // The number of nodes in a tree; allows weighted union to work (flatter tree)
/**
* Nodes (zero based) are initialized as not connected
* @param n Number of nodes
*/
public QuickUnionFind(int n) {
N = n;
id = new int[N];
sz = new int[N];
/* Initialize all nodes to point to themselves.
* At first, all nodes are disconnected
*/
for (int i=0; i<N; i++) {
id[i] = i;
sz[i] = 1; // Each tree has one node
}
}
private int root(int i) {
while (i != id[i]) {
id[i] = id[id[i]]; // Path compression
i = id[i];
}
return i;
}
@Override
public void union(int p, int q) {
int i = root(p);
int j = root(q);
/*
* Here we use weighting to keep the tree flat.
* The smaller tree is merged to the large one
* A flat tree has a faster find root time
*/
if (sz[i] < sz[j]) {
id[i] = j;
sz[j] += sz[i];
} else {
id[j] = i;
sz[i] += sz[j];
}
}
@Override
public boolean connected(int p, int q) {
return root(p) == root(q);
}
@Override
public int getNumberOfTrees() {
Set<Integer> s = new HashSet<Integer>();
for (int i=0; i<N; i++)
s.add(id[i]);
return s.size();
}
}
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