【loj6437】 【PKUSC2018】 PKUSC 计算几何

  1 #include<bits/stdc++.h>
  2 #define M 505
  3 #define DB long double 
  4 #define eps 1e-6
  5 #define PI 3.14159265358979323846
  6 using namespace std;
  7 
  8 bool zero(DB x){return fabs(x)<eps;}
  9 struct pt{
 10     DB x,y;
 11     pt(){x=y=0;}
 12     pt(DB xx,DB yy){x=xx; y=yy;}
 13     friend pt operator -(pt a,pt b){return pt(a.x-b.x,a.y-b.y);} 
 14     friend bool operator <(pt a,pt b){return atan2(a.y,a.x)<atan2(b.y,b.x);}
 15     DB mo(){return x*x+y*y;}
 16 };
 17         
 18 struct line{
 19     DB a,b,c;
 20     line(){a=b=c=0;}
 21     line(DB aa,DB bb,DB cc){a=aa; b=bb; c=cc;}
 22     line(pt A,pt B){
 23         DB x1=A.x,y1=A.y;
 24         DB x2=B.x,y2=B.y;
 25         a=y1-y2; b=x2-x1;
 26         c=-a*x2-b*y2;
 27     }
 28 };
 29 
 30 pt a[M],b[M],c[M]; int n,m;
 31 int type[M]={0};
 32 DB sita[M]={0};
 33 DB solve(pt hh){
 34     memset(c,0,sizeof(c)); 
 35     memset(type,0,sizeof(type));
 36     int cnt=0,ok=1;
 37     DB R=hh.x*hh.x+hh.y*hh.y,r=sqrt(R);
 38     for(int i=1;i<=m;i++){
 39         line p=line(b[i],b[i+1]);
 40         if(fabs(p.c/sqrt(p.a*p.a+p.b*p.b))-eps<r) ok=0; 
 41         DB xl=min(b[i].x,b[i+1].x),xr=max(b[i].x,b[i+1].x);
 42         DB yl=min(b[i].y,b[i+1].y),yr=max(b[i].y,b[i+1].y);
 43         xl-=eps; yl-=eps; xr+=eps; yr+=eps;
 44         DB x1,x2,y1,y2; bool is0=0;
 45         if(zero(p.a)){
 46             y1=y2=-p.c/p.b;
 47             DB delta=R-y1*y1;
 48             if(delta<-eps) continue;
 49             if(zero(delta)) is0=1;
 50             x1=-sqrt(delta); x2=sqrt(delta);
 51         }else{
 52             DB A=p.b*p.b+p.a*p.a;
 53             DB B=p.b*p.c*2;
 54             DB C=p.c*p.c-R*p.a*p.a;
 55             DB delta=B*B-4*A*C;
 56             if(delta<-eps) continue;
 57             if(zero(delta)) is0=1;
 58             y1=(-B+sqrt(delta))/(2*A);
 59             y2=(-B-sqrt(delta))/(2*A);
 60             x1=(-p.b*y1-p.c)/p.a;
 61             x2=(-p.b*y2-p.c)/p.a;
 62         }
 63         int ok1,ok2;
 64         if(xl<=x1&&x1<=xr&&yl<=y1&&y1<=yr){ c[++cnt]=pt(x1,y1); if(is0) continue;}
 65         if(xl<=x2&&x2<=xr&&yl<=y2&&y2<=yr) c[++cnt]=pt(x2,y2); 
 66     }
 67     if(ok){
 68         //return 1;
 69         c[cnt=1]=pt(0,0);
 70     }else{
 71         sort(c+1,c+cnt+1);
 72         for(int i=1;i<=cnt;i++) sita[i]=atan2(c[i].y,c[i].x);
 73     }
 74     c[cnt+1]=c[1]; sita[cnt+1]=sita[1]+2*PI;
 75     DB ans=0;
 76     for(int i=1;i<=cnt;i++){
 77         double SITA=(sita[i]+sita[i+1])/2; 
 78         pt hh=pt(r*cos(SITA)+eps*2,r*sin(SITA));
 79         int cnt=0;
 80         for(int j=1;j<=m;j++){
 81             double l=b[j].x,r=b[j+1].x;
 82             if(zero(b[j].x-b[j+1].x)) continue; 
 83             line p=line(b[j],b[j+1]);
 84             double y=-(p.a*hh.x+p.c)/p.b; 
 85             if(l>r) swap(l,r);
 86             l+=eps; r-=eps;
 87             if(l<=hh.x&&hh.x<=r&&y+eps>hh.y) cnt++;
 88         }
 89         if(cnt&1) ans+=sita[i+1]-sita[i];
 90     }
 91     ans/=(2.*PI);
 92     return ans;
 93 }
 94         
 95 
 96 int main(){
 97 //    freopen("in.txt","r",stdin);
 98 //    freopen("out.txt","w",stdout);
 99     scanf("%d%d",&n,&m);
100     for(int i=1;i<=n;i++) cin>>a[i].x>>a[i].y;
101     for(int i=1;i<=m;i++) cin>>b[i].x>>b[i].y;
102     b[m+1]=b[1];
103     DB ans=0;
104     for(int i=1;i<=n;i++){
105         ans+=solve(a[i]);
106         //printf("%.5Lf\n",ans);
107     }
108     double res=ans;
109     printf("%.5lf\n",res);
110 }