Gym - 101208J 2013 ACM-ICPC World Finals J.Pollution Solution 圆与多边形面积交

题面

题意:给你一个半圆,和另一个多边形(可凹可凸),求面积交

题解:直接上板子,因为其实这个多边形不会穿过这个半圆,所以他和圆的交也就是和半圆的交

打的时候队友说凹的不行,不是板题,后面想想,圆与多边形面积交本来就是拆成有向三角形做的,所以无论凹凸了

 #include<bits/stdc++.h>
 #define inf 1000000000000
 #define M 100009
 #define eps 1e-12
 #define PI acos(-1.0)
 using namespace std;
 struct Point
 {
     double x,y;
     Point(){}
     Point(double xx,double yy){x=xx;y=yy;}
     Point operator -(Point s){return Point(x-s.x,y-s.y);}
     Point operator +(Point s){return Point(x+s.x,y+s.y);}
     double operator *(Point s){return x*s.x+y*s.y;}
     double operator ^(Point s){return x*s.y-y*s.x;}
 }p[M];
 double max(double a,double b){return a>b?a:b;}
 double min(double a,double b){return a<b?a:b;}
 double len(Point a){return sqrt(a*a);}
 double dis(Point a,Point b){return len(b-a);}//两点之间的距离
 double cross(Point a,Point b,Point c)//叉乘
 {
     return (b-a)^(c-a);
 }
 double dot(Point a,Point b,Point c)//点乘
 {
     return (b-a)*(c-a);
 }
 int judge(Point a,Point b,Point c)//判断c是否在ab线段上（前提是c在直线ab上）
 {
     if (c.x>=min(a.x,b.x)
        &&c.x<=max(a.x,b.x)
        &&c.y>=min(a.y,b.y)
        &&c.y<=max(a.y,b.y)) return ;
     return ;
 }
 double area(Point b,Point c,double r)
 {
     Point a(0.0,0.0);
     if(dis(b,c)<eps) return 0.0;
     double h=fabs(cross(a,b,c))/dis(b,c);
     if(dis(a,b)>r-eps&&dis(a,c)>r-eps)//两个端点都在圆的外面则分为两种情况
     {
         double angle=acos(dot(a,b,c)/dis(a,b)/dis(a,c));
         if(h>r-eps) return 0.5*r*r*angle;else
         if(dot(b,a,c)>&&dot(c,a,b)>)
         {
             double angle1=*acos(h/r);
             return 0.5*r*r*fabs(angle-angle1)+0.5*r*r*sin(angle1);
         }else return 0.5*r*r*angle;
     }else
         if(dis(a,b)<r+eps&&dis(a,c)<r+eps) return 0.5*fabs(cross(a,b,c));//两个端点都在圆内的情况
         else//一个端点在圆上一个端点在圆内的情况
         {
             if(dis(a,b)>dis(a,c)) swap(b,c);//默认b在圆内
             if(fabs(dis(a,b))<eps) return 0.0;//ab距离为0直接返回0
             if(dot(b,a,c)<eps)
             {
                 double angle1=acos(h/dis(a,b));
                 double angle2=acos(h/r)-angle1;
                 double angle3=acos(h/dis(a,c))-acos(h/r);
                 return 0.5*dis(a,b)*r*sin(angle2)+0.5*r*r*angle3;
             }else
             {
                 double angle1=acos(h/dis(a,b));
                 double angle2=acos(h/r);
                 double angle3=acos(h/dis(a,c))-angle2;
                 return 0.5*r*dis(a,b)*sin(angle1+angle2)+0.5*r*r*angle3;
             }
         }
 }
 int main()
 {
     int n;
     double R;
     scanf("%d%lf",&n,&R);
     for(int i=;i<=n;i++) scanf("%lf%lf",&p[i].x,&p[i].y);
     p[n+]=p[];
     Point O(,);
     for(int i=;i<=n+;i++) p[i]=p[i]-O;
     O=Point(,);
     double sum=;
     for(int i=;i<=n;i++)
     {
         int j=i+;
         double s=area(p[i],p[j],R);
         if(cross(O,p[i],p[j])>) sum+=s; else sum-=s;
     }
     printf("%.12lf\n",fabs(sum));
     return ;
 }