Gym-100935I Farm 计算几何 圆和矩形面积交

题面

题意:就是给你一个圆,和你一个矩形,求面积并,且 保证是一种情况:三角剖分后 一个点在圆内 两个在圆外

题解:可以直接上圆与凸多边形交的板子,也可以由这题实际情况,面积等于扇形减两个三角形

 #include<bits/stdc++.h>
 using namespace std;
 int main()
 {
     int T;
     double dx,dy,yx,ux,uy,rx,ry,R;
     double s1,s2,s3,x1,y1,x2,y2,cosA;
     scanf("%d",&T);
     for (int i=;i<=T;i++)
     {
         scanf("%lf%lf%lf",&rx,&ry,&R);
         scanf("%lf%lf%lf%lf",&dx,&dy,&ux,&uy);
         dx-=rx;ux-=rx;dy-=ry;uy-=ry;
         x1=dx;
         y1=-sqrt(R*R-x1*x1);
         y2=uy;
         x2=sqrt(R*R-y2*y2);
         cosA=acos((x1*x2+y1*y2)/sqrt((x1*x1+y1*y1)*(x2*x2+y2*y2)));
         s1=R*R*cosA/;
         s2=x1*(y2-y1)/;
         s3=y2*(x1-x2)/;
         printf("Case %d: %.5lf\n",i,s1-s2-s3);
     }
 }

 #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 T,n=;
     double rx,ry,R;
     scanf("%d",&T);
     for (int ii=;ii<=T;ii++)
     {
         scanf("%lf%lf%lf",&rx,&ry,&R);
         scanf("%lf%lf%lf%lf",&p[].x,&p[].y,&p[].x,&p[].y);
         p[].x=p[].x;p[].y=p[].y;
         p[].x=p[].x;p[].y=p[].y;
         p[]=p[];
         Point O(rx,ry);
         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("Case %d: %.5lf\n",ii,fabs(sum));
     }
     return ;
 }