我有一个矩形和一个圆形。我需要验证一个矩形是否在该圆圈内。
我尝试了 Shape.intersects 但 intersects 被检查为数字 1。
有人知道javafx中的这种算法吗？
举例来说，图中只有矩形1、2、3、4在圆内。
感谢您的帮助。
【问题讨论】：
这个解决方案背后的基本思想是，任何多边形都包含在任何凸（见 cmets）形状中，只要多边形内的每个点都在该形状内。如果多边形的至少一个点在形状内，您尝试使用的intersects() 方法将返回true。您已经发现它会起作用，但它也会为任何部分相交的形状提供误报。为了解决这个问题，我们定义了自己的交叉测试，它会查看所有点。
这可以概括为扫描任何给定多边形以查找具有任何给定形状的“总交点”：
public boolean totalIntersects(Polygon poly, Shape testShape) {
List<Point> points = flatDoublesToPoints(poly.getPoints());
boolean inside = true; // If this is false after testing all points, the poly has at least one point outside of the shape.
for(Point point : points) {
if(!testShape.intersects(point.x, point.y, 1, 1)) { // The 3rd and 4th parameters here are "width" and "height". 1 for a point.
inside = false;
}
}
return inside;
}
其中flatDoublesToPoints() 和Point 定义为：
private List<Point> flatDoublesToPoints(List<Double> flatDoubles) {
List<Point> points = new ArrayList<>();
for(int i = 0; i < flatDoubles.size(); i += 2) {
points.add(new Point(flatDoubles.get(i), flatDoubles.get(i + 1)));
}
return points;
}
class Point {
public double x, y;
public Point(double x, double y) {
this.x = x;
this.y = y;
}
}
flatDoublesToPoints() 用于将“平面”{x1, y1, x2, y2, x3, y3...} 多边形列表拆分为更易于理解的数据结构。但是，如果您要进行大量比较，则跳过此步骤可能会有所帮助，出于内存原因直接对“平面列表”进行操作。
以下将其他方法应用于与您的情况极为相似的情况。 （不准确，因为我没有你的代码。）
public class Main extends Application {
public static final int SIZE = 600;
@Override
public void start(Stage primaryStage) throws Exception {
Pane rootPane = new Pane();
List<Rectangle> rects = new ArrayList<>();
for (int j = 0; j < 2; j++) {
for(int i = 0; i < 5; i++) {
Rectangle r = new Rectangle(i * 100, j == 0 ? 0 : 300, 100, 200);
r.setFill(Color.BEIGE);
r.setStroke(Color.BLACK);
rects.add(r);
}
}
rootPane.getChildren().addAll(rects);
Circle circle = new Circle(350, 100, 200);
circle.setStroke(Color.BLACK);
circle.setFill(null);
rootPane.getChildren().add(circle);
List<Polygon> polys = new ArrayList<>();
for(Rectangle rect : rects) {
polys.add(rectangleToPolygon(rect));
}
List<Polygon> intersects = getTotalIntersections(polys, circle);
System.out.println(intersects);
primaryStage.setScene(new Scene(rootPane, SIZE, SIZE));
primaryStage.show();
}
public List<Polygon> getTotalIntersections(List<Polygon> polys, Shape testShape) {
List<Polygon> intersections = new ArrayList<>();
for(Polygon poly : polys) {
if(totalIntersects(poly, testShape)) {
intersections.add(poly);
}
}
return intersections;
}
public static Polygon rectangleToPolygon(Rectangle rect) {
double[] points = {rect.getX(), rect.getY(),
rect.getX() + rect.getWidth(), rect.getY(),
rect.getX() + rect.getWidth(), rect.getY() + rect.getHeight(),
rect.getX(), rect.getY() + rect.getHeight()};
return new Polygon(points);
}
public static void main(String[] args) {
Main.launch(args);
}
}
此代码将打印以下内容：
[Polygon[points=[200.0, 0.0, 300.0, 0.0, 300.0, 200.0, 200.0, 200.0], fill=0x000000ff], Polygon[points=[300.0, 0.0, 400.0, 0.0, 400.0, 200.0, 300.0, 200.0], fill=0x000000ff], Polygon[points=[400.0, 0.0, 500.0, 0.0, 500.0, 200.0, 400.0, 200.0], fill=0x000000ff]]
标记为 2、3 和 4 的三个多边形是哪一个。
【讨论】：
我认为JavaFX不会针对这种情况有一些特殊的方法。
要绘制该圆，您需要中心坐标(X_c, Y_c) 和半径(R)。
要绘制矩形，您需要有角度点的坐标（(X_1, Y_1)、(X_2, Y_2) 等）。
那么你只需要检查矩形的所有点是否都在圆内：
(X_1 - X_c)^2 + (Y_1 - Y_c)^2 < R^2
(X_2 - X_c)^2 + (Y_2 - Y_c)^2 < R^2
...
【讨论】：
X 和Y。假设左上点坐标为(X_1, Y_1)，右上为(X_2, Y_2)等
试试这个：
import javafx.geometry.Point2D;
import javafx.scene.shape.Circle;
import javafx.scene.shape.Rectangle;
/*
Check if a rectangle is contained with in a circle by checking
all rectangle corners.
For the rectangle to be contained in a circle, all its corners should be
in a distance smaller or equal to the circle's radius, from the circle's center.
Note:
Requires some more testing. I tested only a few test cases.
I am not familiar with javafx. This solution does not take into
calculation rectangle's arc or other attributes I may not be aware of.
*/
public class Test{
//apply
public static void main(String[] args){
Circle circle = new Circle(0 ,0, 100);
Rectangle rec = new Rectangle(0, 0, 50 , 50);
System.out.println("Is rectungle inside the circle ? "
+ isContained(circle,rec));
}
//check if rectangle is contained within a circle
private static boolean isContained(Circle circle,Rectangle rec) {
boolean isInside = true;
//get circle center & radius
Point2D center = new Point2D(circle.getCenterX(), circle.getCenterY());
double radius= circle.getRadius();
Point2D[] corners = getRectangleCorners(rec);
for(Point2D corner : corners) {
//if any corner falls outside the circle
//the rectangle is not contained in the circle
if(distanceBetween2Points(corner, center) > radius) {
return false;
}
}
return isInside;
}
//calculate distance between two points
//(updated a per fabian's suggestion)
private static double distanceBetween2Points
(Point2D corner, Point2D center) {
return corner.distance(center);
}
private static Point2D[] getRectangleCorners(Rectangle rec) {
Point2D[] corners = new Point2D[4];
corners[0] = new Point2D(rec.getX(), rec.getY());
corners[1] = new Point2D(rec.getX()+ rec.getWidth() , rec.getY());
corners[2] = new Point2D(rec.getX()+ rec.getWidth(), rec.getY()+ rec.getHeight());
corners[3] = new Point2D(rec.getX(), rec.getY()+ rec.getHeight());
return corners;
}
}
【讨论】：
import javafx.geometry.Point2D;的导入则不会@
这里有一个简单的解决方案：https://stackoverflow.com/a/8721483/1529139
复制粘贴到这里：
class Boundary {
private final Point[] points; // Points making up the boundary
...
/**
* Return true if the given point is contained inside the boundary.
* See: http://www.ecse.rpi.edu/Homepages/wrf/Research/Short_Notes/pnpoly.html
* @param test The point to check
* @return true if the point is inside the boundary, false otherwise
*
*/
public boolean contains(Point test) {
int i;
int j;
boolean result = false;
for (i = 0, j = points.length - 1; i < points.length; j = i++) {
if ((points[i].y > test.y) != (points[j].y > test.y) &&
(test.x < (points[j].x - points[i].x) * (test.y - points[i].y) / (points[j].y-points[i].y) + points[i].x)) {
result = !result;
}
}
return result;
}
}
【讨论】：