半平面交的模板

网上看到个 把半平面交转化为求凸包的方法

也即是把一条直线y=kx+b看成一个点(k,b)求凸包

详细证明及其代码见：http://trinkle.blog.uoj.ac/blog/235

#include <bits/stdc++.h>using namespace std;#define REP(i,a,b)  for(int i=a;i<=(int)(b);++i)#define REPD(i,a,b) for(int i=a;i>=(int)(b);--i)const double PI=acos(-1);const double EPS=1e-6;int dcmp(double x){if(fabs(x)<EPS) return 0;return x > 0 ? 1 : -1;}struct Point{double x,y;};typedef Point Vector;bool operator < (const Point& p1, const Point& p2){return p1.x<p2.x || (p1.x==p2.x && p1.y<p2.y) ;}Vector operator / (const Point& A, double x){return Vector{A.x/x, A.y/x};}Vector operator * (const Vector& A, double x){return Vector{A.x*x, A.y*x};}Vector operator - (const Vector& A, const Vector& B){return Vector{A.x-B.x,A.y-B.y};}Vector operator + (const Point& A, const Vector& v){return Point{A.x+v.x,A.y+v.y};}Vector Rotate(const Point& p,double ang){return Vector{p.x*cos(ang)-p.y*sin(ang),p.x*sin(ang)+p.y*cos(ang)};}double Cross(const Vector& A, const Vector& B){return A.x*B.y-A.y*B.x;}double Dot(Vector A, Vector B){return A.x*B.x+A.y*B.y;}double Length(Vector v){return sqrt(fabs(Dot(v,v)));}Vector Unit(Vector v){return v/Length(v);}Point GetLineIntersection(Point P, Vector v, Point Q, Vector w){Vector u=P-Q;double t=Cross(w,u) / Cross(v,w);return P+v*t;}bool Onsegment(Point p, Point A, Point B){return dcmp(Cross(A-p, B-p)) == 0 && dcmp(Dot(A-p, B-p)) < 0 ;}struct Line{Point p;Vector v;double ang;Line(){}Line(Point p, Vector v):p(p),v(v){ ang=atan2(v.y,v.x); }bool operator < (const Line& l) const {return ang < l.ang;}};typedef vector<Point> Polygon;Polygon CutPolygon(Polygon poly, Point A, Point B){  //A-B切多边形polyPolygon newpoly;int n=poly.size();for(int i=0;i<n;i++){Point C=poly[i];Point D=poly[(i+1)%n];if(dcmp(Cross(B-A, C-A))>0) newpoly.push_back(C);if(dcmp(Cross(B-A, C-D))!=0){   //保证两线不共线Point ip=GetLineIntersection(A,B-A,C,D-C);if(Onsegment(ip,C,D)) newpoly.push_back(ip);}}return newpoly;}bool OnLeft(Point p, Line L){return dcmp(Cross(L.v, p-L.p)) > 0;}Point GetLineIntersection(Line a, Line b){Vector u=a.p-b.p;double t=Cross(b.v, u) / Cross(a.v, b.v);return a.p+a.v*t;}int HalfplaneIntersection(Line* L, int n, Point* poly){  //L组成的半平面交，前提是如果是无界的，那么首先要加个"包围框"sort(L, L+n);int first,last;Line* Q=new Line[n];Point* p=new Point[n];Q[first=last=0]=L[0];for(int i=1;i<n;i++) {while(first<last && !OnLeft(p[last-1], L[i])) last--;   //最后一个点不在L[i]的左侧while(first<last && !OnLeft(p[first], L[i])) first++;   //第一个一个点不在L[i]的左侧Q[++last]=L[i];if(dcmp(Cross(Q[last].v, Q[last-1].v)) == 0){    //最后两条线平行last--;if(OnLeft(L[i].p, Q[last])) Q[last]=L[i];   //新加入的线L[i]在最后一条线的左侧，更新最后一条线}if(first<last) p[last-1]=GetLineIntersection(Q[last-1], Q[last]);}while(first<last && !OnLeft(p[last-1], Q[first])) last--;  //确保最后一个点在第一套线的左侧if(last-first<=1) return 0;    //形成不了封闭的图形  p[last]=GetLineIntersection(Q[first], Q[last]);    //第一条线和最后一条线求交点int m=0;for(int i=first;i<=last;i++) poly[m++]=p[i];    //把答案添入polyreturn m;}