我需要一个算法来检测一个圆是否击中了一个正方形,我看到了这个帖子: Circle-Rectangle collision detection (intersection)
看起来我应该选择ShreevatsaR的答案,但我是个数学傻瓜,而且我不知道如何完成算法。有没有人能找到时间为我做一个完整的例子,我已经在网上搜索了这个,但还没有找到合适的例子。
非常感谢你 索伦
编辑:
好的,这是我的尝试。它不起作用,它从不检测任何碰撞。
typedef struct {
double x;
double y;
} point;
typedef struct {
point one;
point two;
} segment;
typedef struct {
point center;
double radius;
} circle;
typedef struct {
point p;
int width;
int height;
point a;
point b;
point c;
point d;
} rectangle;
double slope(point one, point two) {
return (double)(one.y-two.y)/(one.x-two.x);
}
double distance(point p, segment s) {
// Line one is the original line that was specified, and line two is
// the line we're constructing that runs through the specified point,
// at a right angle to line one.
//
// if it's a vertical line return the horizontal distance
if ( s.one.x == s.two.x)
return fabs(s.one.x - p.x);
// if it's a horizontal line return the vertical distance
if ( s.one.y == s.two.y )
return fabs(s.one.y - p.y);
// otherwise, find the slope of the line
double m_one = slope(s.one, s.two);
// the other slope is at a right angle.
double m_two = -1.0 / m_one;
// find the y-intercepts.
double b_one = s.one.y - s.one.x * m_one;
double b_two = p.y - p.x * m_two;
// find the point of intersection
double x = (b_two - b_one) / (m_one - m_two);
double y = m_one * x + b_one;
// find the x and y distances
double x_dist = x - p.x;
double y_dist = y - p.y;
// and return the total distance.
return sqrt(x_dist * x_dist + y_dist * y_dist);
}
bool intersectsCircle(segment s, circle c) {
return distance(c.center, s) <= c.radius;
}
bool pointInRectangle(point p, rectangle r)
{
float right = r.p.x + r.width;
float left = r.p.x - r.width;
float top = r.p.y + r.height;
float bottom = r.p.y - r.height;
return ((left <= p.x && p.x <= right) && (top <= p.y && p.y <= bottom));
}
bool intersect(circle c, rectangle r) {
segment ab;
ab.one = r.a;
ab.two = r.b;
segment bc;
ab.one = r.b;
ab.two = r.c;
segment cd;
ab.one = r.c;
ab.two = r.d;
segment da;
ab.one = r.d;
ab.two = r.a;
return pointInRectangle(c.center, r) ||
intersectsCircle(ab, c) ||
intersectsCircle(bc, c) ||
intersectsCircle(cd, c) ||
intersectsCircle(da, c);
}
答案 0 :(得分:1)
他似乎留下的主要部分是InteresectsCircle(直线,圆圈)。
#include <math.h>
typedef struct {
double x;
double y;
} point;
typedef struct {
point one;
point two;
} segment;
typedef struct {
point center;
double radius;
} circle;
double slope(point &one, point &two) {
return (double)(one.y-two.y)/(one.x-two.x);
}
double distance(point &p, segment &s) {
// Line one is the original line that was specified, and line two is
// the line we're constructing that runs through the specified point,
// at a right angle to line one.
//
// if it's a vertical line return the horizontal distance
if ( s.one.x == s.two.x)
return fabs(s.one.x - p.x);
// if it's a horizontal line return the vertical distance
if ( s.one.y == s.two.y )
return fabs(s.one.y - p.y);
// otherwise, find the slope of the line
double m_one = slope(s.one, s.two);
// the other slope is at a right angle.
double m_two = -1.0 / m_one;
// find the y-intercepts.
double b_one = s.one.y - s.one.x * m_one;
double b_two = p.y - p.x * m_two;
// find the point of intersection
double x = (b_two - b_one) / (m_one - m_two);
double y = m_one * x + b_one;
// find the x and y distances
double x_dist = x - p.x;
double y_dist = y - p.y;
// and return the total distance.
return sqrt(x_dist * x_dist + y_dist * y_dist);
}
bool IntersectsCircle(segment s, circle c) {
return distance(circle.center, s) <= circle.radius;
}
答案 1 :(得分:1)
我有some code in C++(轻度模板化)应该进行这些交叉测试,但我还没有时间测试它们。特别是,我有段圆交叉测试以及平行四边形圆交叉,它应该计算交叉区域和交叉点。同样,在撰写本评论时,这是完全未经测试的,因此您需要根据自己的需要对其进行测试/调整。