计算几何之求圆与直线的交点
作者:互联网
求圆与直线的交点的方法是:
- 求圆心c在直线l上的投影点pr
- 求出直线l上的单位向量e
- 根据r和pr的长度来计算出圆内线段部分的一半base
- 用pr±base*e即得到答案
题目:CGL_7_D
AC代码:
#include <iostream>
#include <cstdio>
#include <math.h>
using namespace std;
#define COUNTER_CLOCKWISE -1 //逆时针
#define CLOCKWISE 1 //顺时针
#define ONLINE_BACK -2 //p2 p0 p1依次排列在一条直线上
#define ONLINE_FRONT 2 //p0 p1 p2依次排列在一条直线上
#define ON_SEGMENT 0 //p2在线段p0p1上
#define EPS 1E-8
class Point
{
public:
double x, y;
Point()
{
}
Point(double x, double y)
{
(*this).x = x;
(*this).y = y;
}
double operator^(const Point &p) const //叉乘
{
return x * p.y - y * p.x;
}
double operator*(const Point &p) const //点乘
{
return x * p.x + y * p.y;
}
Point operator*(const double &d) const
{
return Point(x * d, y * d);
}
Point operator/(const double &d) const
{
return Point(x / d, y / d);
}
Point operator-(const Point &p) const
{
return Point(x - p.x, y - p.y);
}
Point operator+(const Point &p) const
{
return Point(x + p.x, y + p.y);
}
double sqr()
{
return x * x + y * y;
}
double abs()
{
return sqrt(sqr());
}
double distance(const Point &p)
{
return fabs((*this - p).abs());
}
void print()
{
printf("%.8lf %.8lf", x, y);
}
};
class Line
{
public:
Point p1, p2;
Point vec;
Line()
{
}
Line(Point p1, Point p2)
{
(*this).p1 = p1;
(*this).p2 = p2;
(*this).vec = p2 - p1;
}
};
class Circle
{
public:
Point o;
double r;
Circle()
{
}
Circle(Point o, double r)
{
(*this).o = o;
(*this).r = r;
}
Circle(double cx, double cy, double r)
{
(*this).o.x = cx;
(*this).o.y = cy;
(*this).r = r;
}
};
double get_distance_LP(const Line &line, const Point &p) //点到直线的距离
{
return fabs((p - line.p1) ^ (line.p1 - line.p2) / (line.p1 - line.p2).abs());
}
pair<Point, Point> get_Cross_Points(Circle c, Line l)
{
//l.p1.print();
Point pr = l.p1 + (l.vec) * ((c.o - l.p1) * l.vec)/ ( l.vec.sqr());
double base = sqrt(c.r * c.r - (pr-c.o).sqr());
if (base <= EPS) //只有一个交点
{
return make_pair(pr, pr);
}
//有两个交点
Point e = (l.vec) / l.vec.abs();
return make_pair(pr + e * base, pr - e * base);
}
int main()
{
double cx, cy, r;
cin >> cx >> cy >> r;
Circle c(cx, cy, r);
int q;
cin >> q;
while (q--)
{
Line l = Line();
cin >> l.p1.x >> l.p1.y >> l.p2.x >> l.p2.y;
l.vec = l.p2-l.p1;
auto ans = get_Cross_Points(c, l);
//ans.first.print();
if (ans.first.x < ans.second.x )
{
ans.first.print();
cout << ' ';
ans.second.print();
cout << endl;
}
else if (ans.second.x < ans.first.x)
{
ans.second.print();
cout << ' ';
ans.first.print();
cout << endl;
}
else
{
if (ans.first.y - ans.second.y <= EPS)
{
ans.first.print();
cout << ' ';
ans.second.print();
cout << endl;
}
else
{
ans.second.print();
cout << ' ';
ans.first.print();
cout << endl;
}
}
}
}
转载请注明来源:https://www.longjin666.top/?p=793
标签:p2,直线,p1,const,之求,double,Point,交点,return 来源: https://blog.csdn.net/qq_34026204/article/details/113831971