#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std ;
const double eps = 1e-8;
const double PI = acos(-1.0);
int sgn(double x)
{
if(fabs(x) < eps)return 0;
if(x < 0)return -1;
else return 1;
}
struct Point
{
double x,y;
Point(){}
Point(double _x,double _y)
{
x = _x;y = _y;
}
Point operator -(const Point &b)const
{
return Point(x - b.x,y - b.y);
}
//叉积
double operator ^(const Point &b)const
{
return x*b.y - y*b.x;
}
//点积
double operator *(const Point &b)const
{
return x*b.x + y*b.y;
}
//绕原点旋转角度B（弧度值），后x,y的变化
void transXY(double B)
{
double tx = x,ty = y;
x = tx*cos(B) - ty*sin(B);
y = tx*sin(B) + ty*cos(B);
}
};
struct Line
{
Point s,e;
Line(){}
Line(Point _s,Point _e)
{
s = _s;e = _e;
}
//两直线相交求交点
//第一个值为0表示直线重合，为1表示平行，为0表示相交,为2是相交
//只有第一个值为2时，交点才有意义
pair<int,Point> operator &(const Line &b)const
{
Point res = s;
if(sgn((s-e)^(b.s-b.e)) == 0)
{
if(sgn((s-b.e)^(b.s-b.e)) == 0)
return make_pair(0,res);//重合
else return make_pair(1,res);//平行
}
double t = ((s-b.s)^(b.s-b.e))/((s-e)^(b.s-b.e));
res.x += (e.x-s.x)*t;
res.y += (e.y-s.y)*t;
return make_pair(2,res);
}
};
//*两点间距离
double dist(Point a,Point b)
{
return sqrt((a-b)*(a-b));
}
//*判断三点共线
bool online(Point p1, Point p2, Point p3)
{
return
sgn(p3.x-min(p1.x,p2.x)) >= 0 &&
sgn(p3.x-max(p1.x,p2.x)) <= 0 &&
sgn(p3.y-min(p1.y,p2.y)) >= 0 &&
sgn(p3.y-max(p1.y,p2.y)) <= 0;
}
//*判断线段相交
bool inter(Line l1,Line l2)
{
return
max(l1.s.x,l1.e.x) >= min(l2.s.x,l2.e.x) &&
max(l2.s.x,l2.e.x) >= min(l1.s.x,l1.e.x) &&
max(l1.s.y,l1.e.y) >= min(l2.s.y,l2.e.y) &&
max(l2.s.y,l2.e.y) >= min(l1.s.y,l1.e.y) &&
sgn((l2.s-l1.e)^(l1.s-l1.e))*sgn((l2.e-l1.e)^(l1.s-l1.e)) <= 0 &&
sgn((l1.s-l2.e)^(l2.s-l2.e))*sgn((l1.e-l2.e)^(l2.s-l2.e)) <= 0;
}
//判断直线和线段相交
bool Seg_inter_line(Line l1,Line l2) //判断直线l1和线段l2是否相交
{
return sgn((l2.s-l1.e)^(l1.s-l1.e))*sgn((l2.e-l1.e)^(l1.s-l1.e)) <= 0;
}
//点到直线距离
//返回为result,是点到直线最近的点
Point PointToLine(Point P,Line L)
{
Point result;
double t = ((P-L.s)*(L.e-L.s))/((L.e-L.s)*(L.e-L.s));
result.x = L.s.x + (L.e.x-L.s.x)*t;
result.y = L.s.y + (L.e.y-L.s.y)*t;
return result;
}
//点到线段的距离
//返回点到线段最近的点
Point NearestPointToLineSeg(Point P,Line L)
{
Point result;
double t = ((P-L.s)*(L.e-L.s))/((L.e-L.s)*(L.e-L.s));
if(t >= 0 && t <= 1)
{
result.x = L.s.x + (L.e.x - L.s.x)*t;
result.y = L.s.y + (L.e.y - L.s.y)*t;
}
else
{
if(dist(P,L.s) < dist(P,L.e))
result = L.s;
else result = L.e;
}
return result;
}
//计算多边形面积
//点的编号从0~n-1
double CalcArea(Point p[],int n)
{
double res = 0;
for(int i = 0;i < n;i++)
res += (p[i]^p[(i+1)%n])/2;
return fabs(res);
}
//*判断点在线段上
bool OnSeg(Point P,Line L)
{
return
sgn((L.s-P)^(L.e-P)) == 0 &&
sgn((P.x - L.s.x) * (P.x - L.e.x)) <= 0 &&
sgn((P.y - L.s.y) * (P.y - L.e.y)) <= 0;
}
//*判断点在凸多边形内
//点形成一个凸包，而且按逆时针排序（如果是顺时针把里面的<0改为>0）
//点的编号:0~n-1
//返回值：
//-1:点在凸多边形外
//0:点在凸多边形边界上
//1:点在凸多边形内
int inConvexPoly(Point a,Point p[],int n)
{
for(int i = 0;i < n;i++)
{
if(sgn((p[i]-a)^(p[(i+1)%n]-a)) < 0)return -1;
else if(OnSeg(a,Line(p[i],p[(i+1)%n])))return 0;
}
return 1;
}
//*判断点在任意多边形内
//射线法，poly[]的顶点数要大于等于3,点的编号0~n-1
//返回值
//-1:点在凸多边形外
//0:点在凸多边形边界上
//1:点在凸多边形内
int inPoly(Point p,Point poly[],int n)
{
int cnt;
Line ray,side;
cnt = 0;
ray.s = p;
ray.e.y = p.y;
ray.e.x = -100000000000.0;//-INF,注意取值防止越界
for(int i = 0;i < n;i++)
{
side.s = poly[i];
side.e = poly[(i+1)%n];
if(OnSeg(p,side))return 0;
//如果平行轴则不考虑
if(sgn(side.s.y - side.e.y) == 0)
continue;
if(OnSeg(side.s,ray))
{
if(sgn(side.s.y - side.e.y) > 0)cnt++;
}
else if(OnSeg(side.e,ray))
{
if(sgn(side.e.y - side.s.y) > 0)cnt++;
}
else if(inter(ray,side))
cnt++;
}
if(cnt % 2 == 1)return 1;
else return -1;
}
//判断凸多边形
//允许共线边
//点可以是顺时针给出也可以是逆时针给出
//点的编号0~n-1
bool isconvex(Point poly[],int n)
{
bool s[3];
memset(s,false,sizeof(s));
for(int i = 0;i < n;i++)
{
s[sgn( (poly[(i+1)%n]-poly[i])^(poly[(i+2)%n]-poly[i]) )+1] = true;
if(s[0] && s[2])return false;
}
return true;
}
//过三点求圆心坐标
Point waixin(Point a,Point b,Point c)
{
double a1 = b.x - a.x, b1 = b.y - a.y, c1 = (a1*a1 + b1*b1)/2;
double a2 = c.x - a.x, b2 = c.y - a.y, c2 = (a2*a2 + b2*b2)/2;
double d = a1*b2 - a2*b1;
return Point(a.x + (c1*b2 - c2*b1)/d, a.y + (a1*c2 -a2*c1)/d);
}
int main()
{
return 0 ;
}
