#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 ; }