這裏是另一個https://stackoverflow.com/help/mcve（我在〜3本 「昨天」 開始：00日上午，顯然安德魯·湯普森是在不同的時區;-)）
的基本思路這裏是如下：
這兩個給定的點定義了一條線。也就是說，一條無限的線，而不僅是一條線分段。物體邊界框的角點投影在這條線和它的垂線上。這給出了沿着這些線條的對象範圍（的上限）。這些上限可以用來定義覆蓋物體相應一半所需的線上下的「最小半空間」。這些半空間然後可以與對象相交以獲得期望的結果。
本例中的split方法收到Graphics2D參數。這是用於「調試」的只有 - 也就是說，顯示計算的中間結果（範圍，半空間）以及最終結果的預覽。這個Graphics g參數（和相應的調試輸出）可以簡單地刪除（但它也可能有助於顯示方法的想法）。
import java.awt.Color;
import java.awt.Graphics;
import java.awt.Graphics2D;
import java.awt.Shape;
import java.awt.event.MouseEvent;
import java.awt.event.MouseMotionListener;
import java.awt.geom.AffineTransform;
import java.awt.geom.Area;
import java.awt.geom.Ellipse2D;
import java.awt.geom.Line2D;
import java.awt.geom.Path2D;
import java.awt.geom.Point2D;
import java.awt.geom.Rectangle2D;
import javax.swing.JFrame;
import javax.swing.JPanel;
import javax.swing.SwingUtilities;
public class ShapeSplit
{
public static void main(String[] args)
{
SwingUtilities.invokeLater(new Runnable()
{
@Override
public void run()
{
createAndShowGUI();
}
});
}
private static void createAndShowGUI()
{
JFrame f = new JFrame();
f.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
f.getContentPane().add(new ShapeSplitPanel());
f.setSize(1100,600);
f.setLocationRelativeTo(null);
f.setVisible(true);
}
}
class ShapeSplitPanel extends JPanel implements MouseMotionListener
{
private Shape inputShape = new Ellipse2D.Double(300,200,200,300);
private Point2D point0 = new Point2D.Double(200,300);
private Point2D point1 = new Point2D.Double(600,400);
ShapeSplitPanel()
{
addMouseMotionListener(this);
}
@Override
protected void paintComponent(Graphics gr)
{
super.paintComponent(gr);
Graphics2D g = (Graphics2D)gr;
g.setColor(Color.BLUE);
g.fill(inputShape);
g.setColor(Color.BLACK);
g.draw(new Line2D.Double(point0, point1));
g.fill(new Ellipse2D.Double(
point0.getX() - 3, point0.getY()-3, 6, 6));
g.fill(new Ellipse2D.Double(
point1.getX() - 3, point1.getY()-3, 6, 6));
split(new Area(inputShape), point0, point1, g);
}
private static Area[] split(Area a, Point2D p0, Point2D p1, Graphics2D g)
{
// Compute the direction of the line (L)
// and its perpendicular (P)
double dx = p1.getX() - p0.getX();
double dy = p1.getY() - p0.getY();
double length = Math.hypot(dx, dy);
double dirLx = dx/length;
double dirLy = dy/length;
double dirPx = -dirLy;
double dirPy = dirLx;
// Compute the minimum and maximum of all dot
// products that describe the distance of the
// projection of the corner points of the
// bounding box on on the line (L) and its
// perpendicular (P). These are upper limits
// for the extents of the object along these
// directions
double minDotL = Double.MAX_VALUE;
double maxDotL = -Double.MAX_VALUE;
double minDotP = Double.MAX_VALUE;
double maxDotP = -Double.MAX_VALUE;
Rectangle2D bounds = a.getBounds2D();
for (int i=0; i<4; i++)
{
Point2D corner = getCorner(bounds, i);
double pdx = corner.getX() - p0.getX();
double pdy = corner.getY() - p0.getY();
double dotL = dirLx * pdx + dirLy * pdy;
minDotL = Math.min(minDotL, dotL);
maxDotL = Math.max(maxDotL, dotL);
double dotP = dirPx * pdx + dirPy * pdy;
minDotP = Math.min(minDotP, dotP);
maxDotP = Math.max(maxDotP, dotP);
}
// Compute the start- and end points of
// the line segments describing the
// extent of the bounds along the line
// and the perpendicular
Point2D extentLmin = new Point2D.Double(
p0.getX() + minDotL * dirLx,
p0.getY() + minDotL * dirLy);
Point2D extentLmax = new Point2D.Double(
p0.getX() + maxDotL * dirLx,
p0.getY() + maxDotL * dirLy);
Point2D extentPmin = new Point2D.Double(
p0.getX() + minDotP * dirPx,
p0.getY() + minDotP * dirPy);
Point2D extentPmax = new Point2D.Double(
p0.getX() + maxDotP * dirPx,
p0.getY() + maxDotP * dirPy);
// Compute the two rectangles that cover
// each half of the object based on
// the given line
Path2D half0 = new Path2D.Double();
half0.moveTo(extentLmin.getX(), extentLmin.getY());
half0.lineTo(
extentLmin.getX() + minDotP * dirPx,
extentLmin.getY() + minDotP * dirPy);
half0.lineTo(
extentLmax.getX() + minDotP * dirPx,
extentLmax.getY() + minDotP * dirPy);
half0.lineTo(extentLmax.getX(), extentLmax.getY());
half0.closePath();
Path2D half1 = new Path2D.Double();
half1.moveTo(extentLmin.getX(), extentLmin.getY());
half1.lineTo(
extentLmin.getX() + maxDotP * dirPx,
extentLmin.getY() + maxDotP * dirPy);
half1.lineTo(
extentLmax.getX() + maxDotP * dirPx,
extentLmax.getY() + maxDotP * dirPy);
half1.lineTo(extentLmax.getX(), extentLmax.getY());
half1.closePath();
// Compute the resulting areas by intersecting
// the original area with both halves
Area a0 = new Area(a);
a0.intersect(new Area(half0));
Area a1 = new Area(a);
a1.intersect(new Area(half1));
// Debugging output
if (g != null)
{
g.setColor(Color.GRAY);
g.draw(bounds);
g.setColor(Color.RED);
g.draw(new Line2D.Double(extentLmin, extentLmax));
g.setColor(Color.GREEN);
g.draw(new Line2D.Double(extentPmin, extentPmax));
g.setColor(Color.YELLOW.darker());
g.draw(half0);
g.setColor(Color.MAGENTA);
g.draw(half1);
g.setColor(Color.BLUE);
g.fill(AffineTransform.getTranslateInstance(400, -20).
createTransformedShape(a0));
g.setColor(Color.BLUE);
g.fill(AffineTransform.getTranslateInstance(400, +20).
createTransformedShape(a1));
}
return new Area[] { a0, a1 };
}
private static Point2D getCorner(Rectangle2D r, int corner)
{
switch (corner)
{
case 0: return new Point2D.Double(r.getMinX(), r.getMinY());
case 1: return new Point2D.Double(r.getMinX(), r.getMaxY());
case 2: return new Point2D.Double(r.getMaxX(), r.getMaxY());
case 3: return new Point2D.Double(r.getMaxX(), r.getMinY());
}
return null;
}
@Override
public void mouseDragged(MouseEvent e)
{
point1.setLocation(e.getPoint());
repaint();
}
@Override
public void mouseMoved(MouseEvent e)
{
}
}
編輯一旁：從技術上講，它可以是更容易（或甚至更優雅）來轉換原來的形狀和線，使線與x軸一致，則限定所述半空間（在這種情況下可能是簡單的Rectangle2D s），並將剪輯結果轉換回原始方向。但是我想「就地」計算它，而不必創建許多變形的形狀。
EDIT2：另一個片段的評論，直接插入前// Debugging output
AffineTransform t = new AffineTransform();
double angle = Math.atan2(p1.getY() - p0.getY(), p1.getX() - p0.getX());
t.rotate(-angle, p0.getX(), p0.getY());
a0 = a0.createTransformedArea(t);
a1 = a1.createTransformedArea(t);
EDIT3第二種方法時，只有相關方法此時
private static Area[] split(Area a, Point2D p0, Point2D p1, Graphics2D g)
{
// Compute the angle of the line to the x-axis
double dx = p1.getX() - p0.getX();
double dy = p1.getY() - p0.getY();
double angleRadToX = Math.atan2(dy, dx);
// Align the area so that the line matches the x-axis
AffineTransform at = new AffineTransform();
at.rotate(-angleRadToX);
at.translate(-p0.getX(), -p0.getY());
Area aa = a.createTransformedArea(at);
// Compute the upper and lower halves that the area
// has to be intersected with
Rectangle2D bounds = aa.getBounds2D();
double half0minY = Math.min(0, bounds.getMinY());
double half0maxY = Math.min(0, bounds.getMaxY());
Rectangle2D half0 = new Rectangle2D.Double(
bounds.getX(), half0minY,
bounds.getWidth(), half0maxY-half0minY);
double half1minY = Math.max(0, bounds.getMinY());
double half1maxY = Math.max(0, bounds.getMaxY());
Rectangle2D half1 = new Rectangle2D.Double(
bounds.getX(), half1minY,
bounds.getWidth(), half1maxY-half1minY);
// Compute the resulting areas by intersecting
// the original area with both halves, and
// transform them back to their initial position
Area a0 = new Area(aa);
a0.intersect(new Area(half0));
Area a1 = new Area(aa);
a1.intersect(new Area(half1));
try
{
at.invert();
}
catch (NoninvertibleTransformException e)
{
// Always invertible
}
a0 = a0.createTransformedArea(at);
a1 = a1.createTransformedArea(at);
// Debugging output
if (g != null)
{
g.setColor(Color.GRAY);
g.draw(bounds);
g.setColor(Color.RED);
g.draw(aa);
g.setColor(Color.YELLOW.darker());
g.draw(half0);
g.setColor(Color.MAGENTA);
g.draw(half1);
g.setColor(Color.BLUE.darker());
g.fill(AffineTransform.getTranslateInstance(400, -20).
createTransformedShape(a0));
g.setColor(Color.BLUE.brighter());
g.fill(AffineTransform.getTranslateInstance(400, +20).
createTransformedShape(a1));
}
return new Area[] { a0, a1 };
}
如果我嘗試使用變換方法。我將如何去改變線條和區域，以便旋轉並與x軸對齊？ – Thatdude1
計算線和x軸之間的角度爲Math.atan2（p1.y-p0.y，p1.x-p0.x）'，然後創建一個AffineTransfrom.getRotateInstance（angle，p0.x， p0.y）'，然後用它來創建一個變換後的形狀，然後計算出變換後的形狀的邊界，並將這個邊界矩形分成x軸上下的部分，然後將這些半部分與形狀相交，最後轉換這些使用仿射變換的逆來交叉返回。 – Marco13
更新了我的問題與您的建議..但我的形狀不正確旋轉。根據線的旋轉和平移，有時並不總是旋轉到x軸 – Thatdude1