hdu 5120（求两个圆环相交的面积 2014北京现场赛 I题）

两个圆环的内外径相同 给出内外径 和 两个圆心 求两个圆环相交的面积

画下图可以知道 就是两个大圆交-2*小圆与大圆交+2小圆交

Sample Input
2
2 3
0 0
0 0
2 3
0 0
5 0

Sample Output
Case #1: 15.707963
Case #2: 2.250778

 # include <iostream>
 # include <cstdio>
 # include <cstring>
 # include <algorithm>
 # include <string>
 # include <cmath>
 # include <queue>
 # include <list>
 # define LL long long
 using namespace std ;

 const double eps = 1e-;
 const double PI = acos(-1.0);

 struct Point
 {
     double x,y;
     Point(){}
     Point(double _x,double _y)
     {
         x = _x;y = _y;
     }
     Point operator -(const Point &b)const
     {
         return Point(x - b.x,y - b.y);
     }
     //叉积
     double operator ^(const Point &b)const
     {
         return x*b.y - y*b.x;
     }
     //点积
     double operator *(const Point &b)const
     {
         return x*b.x + y*b.y;
     }
     //绕原点旋转角度B（弧度值），后x,y的变化
     void transXY(double B)
     {
         double tx = x,ty = y;
         x = tx*cos(B) - ty*sin(B);
         y = tx*sin(B) + ty*cos(B);
     }
 };

 //*两点间距离
 double dist(Point a,Point b)
 {
     return sqrt((a-b)*(a-b));
 }

 //两个圆的公共部分面积
 double Area_of_overlap(Point c1,double r1,Point c2,double r2)
 {
     double d = dist(c1,c2);
     if(r1 + r2 < d + eps)return ;
     if(d < fabs(r1 - r2) + eps)
     {
         double r = min(r1,r2);
         return PI*r*r;
     }
     double x = (d*d + r1*r1 - r2*r2)/(*d);
     double t1 = acos(x / r1);
     double t2 = acos((d - x)/r2);
     return r1*r1*t1 + r2*r2*t2 - d*r1*sin(t1);
 }

 int main()
 {
     //freopen("in.txt","r",stdin) ;
     int T ;
     scanf("%d" , &T) ;
     int Case = ;
     while(T--)
     {
         Case++ ;
         double r , R ;
         double x1 , y1 , x2 , y2 ;
         scanf("%lf%lf%lf%lf%lf%lf" , &r , &R , &x1 , &y1 , &x2 , &y2) ;
         Point c1(x1 ,y1) ;
         Point c2(x2 ,y2) ;
         double ans =  ;
         ans = Area_of_overlap(c1,R,c2,R) + Area_of_overlap(c1,r,c2,r);
         ans -= Area_of_overlap(c1,r,c2,R) ;
         ans -= Area_of_overlap(c1,R,c2,r) ;
         printf("Case #%d: %.6lf\n" , Case , ans) ;

     }
     return ;
 }