如何分辨圆和矩形在二维欧几里得空间中是否相交?(即经典二维几何)
当前回答
这里有另一个解决方案,实现起来非常简单(也非常快)。它将捕获所有的交点,包括当球体完全进入矩形时。
// clamp(value, min, max) - limits value to the range min..max
// Find the closest point to the circle within the rectangle
float closestX = clamp(circle.X, rectangle.Left, rectangle.Right);
float closestY = clamp(circle.Y, rectangle.Top, rectangle.Bottom);
// Calculate the distance between the circle's center and this closest point
float distanceX = circle.X - closestX;
float distanceY = circle.Y - closestY;
// If the distance is less than the circle's radius, an intersection occurs
float distanceSquared = (distanceX * distanceX) + (distanceY * distanceY);
return distanceSquared < (circle.Radius * circle.Radius);
任何像样的数学库都可以将其缩短为3或4行。
其他回答
圆与矩形相交只有两种情况:
圆的中心在矩形的内部,或者 矩形的一条边在圆上有一个点。
注意,这并不要求矩形与轴平行。
(一种方法是:如果没有一条边在圆中有点(如果所有的边都完全“在”圆外),那么圆仍然可以与多边形相交的唯一方法是它完全位于多边形内部。)
有了这样的见解,就可以像下面这样工作,其中圆的中心是P,半径是R,矩形的顶点是A, B, C, D(不完整的代码):
def intersect(Circle(P, R), Rectangle(A, B, C, D)):
S = Circle(P, R)
return (pointInRectangle(P, Rectangle(A, B, C, D)) or
intersectCircle(S, (A, B)) or
intersectCircle(S, (B, C)) or
intersectCircle(S, (C, D)) or
intersectCircle(S, (D, A)))
如果你在写任何几何,你的库中可能已经有了上面的函数。否则,pointInRectangle()可以用几种方式实现;任何一般的多边形点方法都可以工作,但对于矩形,你可以检查这是否有效:
0 ≤ AP·AB ≤ AB·AB and 0 ≤ AP·AD ≤ AD·AD
intersectCircle()也很容易实现:一种方法是检查从P到直线的垂线的脚是否足够近并且在端点之间,否则检查端点。
最酷的是,同样的想法不仅适用于矩形,而且适用于一个圆与任何简单多边形的交点——甚至不必是凸多边形!
我为处理形状创建了类 希望你喜欢
public class Geomethry {
public static boolean intersectionCircleAndRectangle(int circleX, int circleY, int circleR, int rectangleX, int rectangleY, int rectangleWidth, int rectangleHeight){
boolean result = false;
float rectHalfWidth = rectangleWidth/2.0f;
float rectHalfHeight = rectangleHeight/2.0f;
float rectCenterX = rectangleX + rectHalfWidth;
float rectCenterY = rectangleY + rectHalfHeight;
float deltax = Math.abs(rectCenterX - circleX);
float deltay = Math.abs(rectCenterY - circleY);
float lengthHypotenuseSqure = deltax*deltax + deltay*deltay;
do{
// check that distance between the centerse is more than the distance between the circumcircle of rectangle and circle
if(lengthHypotenuseSqure > ((rectHalfWidth+circleR)*(rectHalfWidth+circleR) + (rectHalfHeight+circleR)*(rectHalfHeight+circleR))){
//System.out.println("distance between the centerse is more than the distance between the circumcircle of rectangle and circle");
break;
}
// check that distance between the centerse is less than the distance between the inscribed circle
float rectMinHalfSide = Math.min(rectHalfWidth, rectHalfHeight);
if(lengthHypotenuseSqure < ((rectMinHalfSide+circleR)*(rectMinHalfSide+circleR))){
//System.out.println("distance between the centerse is less than the distance between the inscribed circle");
result=true;
break;
}
// check that the squares relate to angles
if((deltax > (rectHalfWidth+circleR)*0.9) && (deltay > (rectHalfHeight+circleR)*0.9)){
//System.out.println("squares relate to angles");
result=true;
}
}while(false);
return result;
}
public static boolean intersectionRectangleAndRectangle(int rectangleX, int rectangleY, int rectangleWidth, int rectangleHeight, int rectangleX2, int rectangleY2, int rectangleWidth2, int rectangleHeight2){
boolean result = false;
float rectHalfWidth = rectangleWidth/2.0f;
float rectHalfHeight = rectangleHeight/2.0f;
float rectHalfWidth2 = rectangleWidth2/2.0f;
float rectHalfHeight2 = rectangleHeight2/2.0f;
float deltax = Math.abs((rectangleX + rectHalfWidth) - (rectangleX2 + rectHalfWidth2));
float deltay = Math.abs((rectangleY + rectHalfHeight) - (rectangleY2 + rectHalfHeight2));
float lengthHypotenuseSqure = deltax*deltax + deltay*deltay;
do{
// check that distance between the centerse is more than the distance between the circumcircle
if(lengthHypotenuseSqure > ((rectHalfWidth+rectHalfWidth2)*(rectHalfWidth+rectHalfWidth2) + (rectHalfHeight+rectHalfHeight2)*(rectHalfHeight+rectHalfHeight2))){
//System.out.println("distance between the centerse is more than the distance between the circumcircle");
break;
}
// check that distance between the centerse is less than the distance between the inscribed circle
float rectMinHalfSide = Math.min(rectHalfWidth, rectHalfHeight);
float rectMinHalfSide2 = Math.min(rectHalfWidth2, rectHalfHeight2);
if(lengthHypotenuseSqure < ((rectMinHalfSide+rectMinHalfSide2)*(rectMinHalfSide+rectMinHalfSide2))){
//System.out.println("distance between the centerse is less than the distance between the inscribed circle");
result=true;
break;
}
// check that the squares relate to angles
if((deltax > (rectHalfWidth+rectHalfWidth2)*0.9) && (deltay > (rectHalfHeight+rectHalfHeight2)*0.9)){
//System.out.println("squares relate to angles");
result=true;
}
}while(false);
return result;
}
}
预检查一个完全封装矩形的圆是否与矩形发生碰撞。 检查圆内的矩形角。 对于每条边,看看是否有一条线与圆相交。将中心点C投影到直线AB上,得到点d。如果CD的长度小于半径,则发生了碰撞。
projectionScalar=dot(AC,AB)/(mag(AC)*mag(AB));
if(projectionScalar>=0 && projectionScalar<=1) {
D=A+AB*projectionScalar;
CD=D-C;
if(mag(CD)<circle.radius){
// there was a collision
}
}
稍微改进一下e。james的回答:
double dx = abs(circle.x - rect.x) - rect.w / 2,
dy = abs(circle.y - rect.y) - rect.h / 2;
if (dx > circle.r || dy > circle.r) { return false; }
if (dx <= 0 || dy <= 0) { return true; }
return (dx * dx + dy * dy <= circle.r * circle.r);
这就减去了一次,而不是最多减去三次。
首先检查矩形和与圆相切的正方形是否重叠(简单)。如果它们不重叠,就不会碰撞。 检查圆的中心是否在矩形内(简单)。如果它在里面,它们就会碰撞。 计算矩形边到圆中心的最小平方距离(略硬)。如果小于半径的平方,它们就会碰撞,否则不会。
它是有效的,因为:
首先,它用一个便宜的算法检查最常见的场景,当它确定它们没有碰撞时,它就结束了。 然后它用一个廉价的算法检查下一个最常见的场景(不要计算平方根,使用平方值),当它确定它们碰撞时,它就结束了。 然后它执行更昂贵的算法来检查与矩形边框的碰撞。