线交叉矩形 - 如何找到交叉点?

时间:2013-03-24 02:15:01

标签: java gwt math vector java-canvas

我在画布上绘制一个矩形,从矩形中心到坐标空间中的某个随机点绘制一条线。

现在,我想根据矩形内部的长度截断线条,使线条从矩形边缘开始。

我怎么能这样做?

示例

  • 矩形可以由2个点定义:Pstart(1, 3)Pend(3, 1)
  • 中心点可以计算为:P(2, 2)
  • 现在绘制从P(2, 2)Q(10, 2)的行。

我知道矩形的宽度是2,我可以告诉该行从P(4, 2)而不是P(2, 2)开始。

当点与XY轴之一不平行时,这会变得更复杂。此外,矩形内部的长度对于对角线而言将是不同的。

如何计算线点相对于矩形中心和线的终点的起始偏移?

可能我必须找到线穿过矩形的点,然后让线从交叉点开始。但我怎么能明白这一点呢?

3 个答案:

答案 0 :(得分:13)

老实说,我不懂数学,但是......

基本上,你有5行。原始线和矩形的4条线。因此,如果你将它分解为一个简单的线路交叉线问题,它应该会变得更容易......

enter image description here

import java.awt.BorderLayout;
import java.awt.Color;
import java.awt.Dimension;
import java.awt.EventQueue;
import java.awt.Graphics;
import java.awt.Graphics2D;
import java.awt.Point;
import java.awt.geom.Ellipse2D;
import java.awt.geom.Line2D;
import java.awt.geom.Point2D;
import java.awt.geom.Rectangle2D;
import javax.swing.JFrame;
import javax.swing.JPanel;
import javax.swing.UIManager;
import javax.swing.UnsupportedLookAndFeelException;

public class IntersectPoint {

    public static void main(String[] args) {
        new IntersectPoint();
    }

    public IntersectPoint() {
        EventQueue.invokeLater(new Runnable() {
            @Override
            public void run() {
                try {
                    UIManager.setLookAndFeel(UIManager.getSystemLookAndFeelClassName());
                } catch (ClassNotFoundException | InstantiationException | IllegalAccessException | UnsupportedLookAndFeelException ex) {
                }

                JFrame frame = new JFrame("Testing");
                frame.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
                frame.setLayout(new BorderLayout());
                frame.add(new TestPane());
                frame.pack();
                frame.setLocationRelativeTo(null);
                frame.setVisible(true);
            }
        });
    }

    public class TestPane extends JPanel {

        @Override
        public Dimension getPreferredSize() {
            return new Dimension(200, 200);
        }

        @Override
        protected void paintComponent(Graphics g) {
            super.paintComponent(g);

            int x = (int) (getWidth() * 0.2f);
            int y = (int) (getHeight() * 0.2f);
            int width = (int) (getWidth() * 0.6f);
            int height = (int) (getHeight() * 0.6f);

            int x1 = x;
            int y1 = 0;
            int x2 = x + width;
            int y2 = getHeight();

            Line2D line = new Line2D.Double(x1, y1, x2, y2);
            Rectangle2D rect = new Rectangle2D.Double(x, y, width, height);

            Graphics2D g2d = (Graphics2D) g.create();
            g2d.draw(rect);
            g2d.draw(line);

            g2d.setColor(Color.RED);
            Point2D[] ps = getIntersectionPoint(line, rect);
            for (Point2D p : ps) {
                if (p != null) {
                    g2d.fill(new Ellipse2D.Double(p.getX() - 4, p.getY() - 4, 8, 8));
                }
            }
            g2d.dispose();

        }

        public Point2D[] getIntersectionPoint(Line2D line, Rectangle2D rectangle) {

            Point2D[] p = new Point2D[4];

            // Top line
            p[0] = getIntersectionPoint(line,
                            new Line2D.Double(
                            rectangle.getX(),
                            rectangle.getY(),
                            rectangle.getX() + rectangle.getWidth(),
                            rectangle.getY()));
            // Bottom line
            p[1] = getIntersectionPoint(line,
                            new Line2D.Double(
                            rectangle.getX(),
                            rectangle.getY() + rectangle.getHeight(),
                            rectangle.getX() + rectangle.getWidth(),
                            rectangle.getY() + rectangle.getHeight()));
            // Left side...
            p[2] = getIntersectionPoint(line,
                            new Line2D.Double(
                            rectangle.getX(),
                            rectangle.getY(),
                            rectangle.getX(),
                            rectangle.getY() + rectangle.getHeight()));
            // Right side
            p[3] = getIntersectionPoint(line,
                            new Line2D.Double(
                            rectangle.getX() + rectangle.getWidth(),
                            rectangle.getY(),
                            rectangle.getX() + rectangle.getWidth(),
                            rectangle.getY() + rectangle.getHeight()));

            return p;

        }

        public Point2D getIntersectionPoint(Line2D lineA, Line2D lineB) {

            double x1 = lineA.getX1();
            double y1 = lineA.getY1();
            double x2 = lineA.getX2();
            double y2 = lineA.getY2();

            double x3 = lineB.getX1();
            double y3 = lineB.getY1();
            double x4 = lineB.getX2();
            double y4 = lineB.getY2();

            Point2D p = null;

            double d = (x1 - x2) * (y3 - y4) - (y1 - y2) * (x3 - x4);
            if (d != 0) {
                double xi = ((x3 - x4) * (x1 * y2 - y1 * x2) - (x1 - x2) * (x3 * y4 - y3 * x4)) / d;
                double yi = ((y3 - y4) * (x1 * y2 - y1 * x2) - (y1 - y2) * (x3 * y4 - y3 * x4)) / d;

                p = new Point2D.Double(xi, yi);

            }
            return p;
        }
    }
}

答案 1 :(得分:3)

查看用于按矩形划线的Liang-Barsky algorithm

答案 2 :(得分:0)

矩形的顶点:a,b,c,d。代表每个人的x和y坐标,如ax,ay等。

该行的端点:x,y

该行跟随y = mx + b,然后向上或向下,向右或向左移动。这会缩小您可能的矩形边缘,以便穿越到2。

使用y = mx + b确定与水平线交叉的垂直坐标,以及与垂直线交叉的水平分量。其中只有一个实际上在你的矩形上(即包含在一个矩形边内),或者它将在一个角落相交。