【半平面交】bzoj2618 [Cqoi2006]凸多边形
#include<cstdio> #include<cmath> #include<algorithm> using namespace std; #define EPS 0.0000001 #define N 511 typedef double db; const db PI=acos(-1.0); struct Point{db x,y;}; typedef Point Vector; Vector operator - (const Point &a,const Point &b){return (Vector){a.x-b.x,a.y-b.y};} Vector operator * (const Vector &a,const db &k){return (Vector){a.x*k,a.y*k};} Vector operator + (const Vector &a,const Vector &b){return (Vector){a.x+b.x,a.y+b.y};} db Cross(Vector a,Vector b){return a.x*b.y-a.y*b.x;} struct Line { Point p; Vector v; db ang; Line(){} Line(const Point &a,const Point &b) { v=b-a; p=a; ang=atan2(v.y,v.x); if(ang<0) ang+=2.0*PI; } }; bool operator < (const Line &a,const Line &b){return a.ang<b.ang;} bool OnLeft(Line l,Point a){return Cross(l.v,a-l.p)>0;} Point GetJiaodian(Line a,Line b){return a.p+a.v*(Cross(b.v,a.p-b.p)/Cross(a.v,b.v));} int n; Point ps[N]; Line q[N]; int head=1,tail=1; Line ls[N]; void BPMJ() { sort(ls+1,ls+n+1); q[1]=ls[1]; for(int i=2;i<=n;++i) { while(head<tail&&(!OnLeft(ls[i],ps[tail-1]))) --tail; while(head<tail&&(!OnLeft(ls[i],ps[head]))) ++head; q[++tail]=ls[i]; if(fabs(Cross(q[tail].v,q[tail-1].v))<EPS) { --tail; if(OnLeft(q[tail],ls[i].p)) q[tail]=ls[i]; } if(head<tail) ps[tail-1]=GetJiaodian(q[tail-1],q[tail]); } while(head<tail&&(!OnLeft(q[head],ps[tail-1]))) --tail; ps[tail]=GetJiaodian(q[tail],q[head]); } int nn,mm; Point tmp[51]; db area; int main() { scanf("%d",&nn); for(int i=1;i<=nn;++i) { scanf("%d",&mm); for(int j=1;j<=mm;++j) scanf("%lf%lf",&tmp[j].x,&tmp[j].y); for(int j=1;j<mm;++j) ls[++n]=Line(tmp[j],tmp[j+1]); ls[++n]=Line(tmp[mm],tmp[1]); } BPMJ(); for(int i=head+1;i<tail;++i) area+=Cross(ps[i]-ps[head],ps[i+1]-ps[head]); printf("%.3lf\n",area/2.0); return 0; }
——The Solution By AutSky_JadeK From UESTC
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