BZOJ - 2618 凸多边形 (半平面交)
题意：求n个凸多边形的交面积。
半平面交模板题。
1 #include<bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 typedef double db; 5 const int N=1000+10,inf=0x3f3f3f3f; 6 const db pi=acos(-1),eps=1e-8; 7 struct P { 8 db x,y; 9 P operator-(P b) {return {x-b.x,y-b.y};} 10 P operator+(P b) {return {x+b.x,y+b.y};} 11 P operator*(db b) {return {x*b,y*b};} 12 bool operator<(const P& b)const {return x!=b.x?x<b.x:y<b.y;} 13 bool operator==(const P& b)const {return x==b.x&&y==b.y;} 14 } p[N],hpi[N]; 15 db cross(P a,P b) {return a.x*b.y-a.y*b.x;} 16 struct Line { 17 P p,v; 18 db rad()const {return atan2(v.y,v.x);}; 19 } line[N],q[N]; 20 bool onleft(P p,Line a) {return cross(a.v,p-a.p)>0;} 21 P its(Line a,Line b) { 22 P v=a.p-b.p; 23 db t=cross(b.v,v)/cross(a.v,b.v); 24 return a.p+a.v*t; 25 } 26 bool operator<(const Line& a,const Line& b) { 27 db r1=a.rad(),r2=b.rad(); 28 return fabs(r1-r2)>eps?r1<r2:onleft(a.p,b); 29 } 30 int n,m,nl,nhpi; 31 void HPI() { 32 sort(line,line+nl); 33 int hd=0,tl=-1; 34 for(int i=0; i<nl; ++i) { 35 if(hd<=tl&&fabs(line[i].rad()-q[tl].rad())<eps)continue; 36 if(hd<tl&&cross(q[hd].v,q[tl].v)<0&&onleft(its(q[hd],q[tl]),line[i]))continue; 37 for(; hd<tl&&!onleft(its(q[tl-1],q[tl]),line[i]); --tl); 38 for(; hd<tl&&!onleft(its(q[hd],q[hd+1]),line[i]); ++hd); 39 q[++tl]=line[i]; 40 } 41 nhpi=0; 42 for(int i=hd; i<tl; ++i)hpi[nhpi++]=its(q[i],q[i+1]); 43 hpi[nhpi++]=its(q[tl],q[hd]); 44 } 45 db area() { 46 db ret=0; 47 for(int i=1; i<nhpi-1; ++i)ret+=cross(hpi[i]-hpi[0],hpi[i+1]-hpi[0]); 48 return fabs(ret/2); 49 } 50 51 int main() { 52 for(scanf("%d",&m); m--;) { 53 scanf("%d",&n); 54 for(int i=0; i<n; ++i)scanf("%lf%lf",&p[i].x,&p[i].y); 55 for(int i=0; i<n; ++i)line[nl++]= {p[i],p[(i+1)%n]-p[i]}; 56 } 57 HPI(); 58 printf("%.3f\n",area()); 59 return 0; 60 }
