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

 1 # include <iostream>
 2 # include <cstdio>
 3 # include <cstring>
 4 # include <algorithm>
 5 # include <string>
 6 # include <cmath>
 7 # include <queue>
 8 # include <list>
 9 # define LL long long
10 using namespace std ;
11 
12 const double eps = 1e-8;
13 const double PI = acos(-1.0);
14 
15 struct Point
16 {
17     double x,y;
18     Point(){}
19     Point(double _x,double _y)
20     {
21         x = _x;y = _y;
22     }
23     Point operator -(const Point &b)const
24     {
25         return Point(x - b.x,y - b.y);
26     }
27     //叉积
28     double operator ^(const Point &b)const
29     {
30         return x*b.y - y*b.x;
31     }
32     //点积
33     double operator *(const Point &b)const
34     {
35         return x*b.x + y*b.y;
36     }
37     //绕原点旋转角度B（弧度值），后x,y的变化
38     void transXY(double B)
39     {
40         double tx = x,ty = y;
41         x = tx*cos(B) - ty*sin(B);
42         y = tx*sin(B) + ty*cos(B);
43     }
44 };
45 
46 //*两点间距离
47 double dist(Point a,Point b)
48 {
49     return sqrt((a-b)*(a-b));
50 }
51 
52 //两个圆的公共部分面积
53 double Area_of_overlap(Point c1,double r1,Point c2,double r2)
54 {
55     double d = dist(c1,c2);
56     if(r1 + r2 < d + eps)return 0;
57     if(d < fabs(r1 - r2) + eps)
58     {
59         double r = min(r1,r2);
60         return PI*r*r;
61     }
62     double x = (d*d + r1*r1 - r2*r2)/(2*d);
63     double t1 = acos(x / r1);
64     double t2 = acos((d - x)/r2);
65     return r1*r1*t1 + r2*r2*t2 - d*r1*sin(t1);
66 }
67 
68 int main()
69 {
70     //freopen("in.txt","r",stdin) ;
71     int T ;
72     scanf("%d" , &T) ;
73     int Case = 0;
74     while(T--)
75     {
76         Case++ ;
77         double r , R ;
78         double x1 , y1 , x2 , y2 ;
79         scanf("%lf%lf%lf%lf%lf%lf" , &r , &R , &x1 , &y1 , &x2 , &y2) ;
80         Point c1(x1 ,y1) ;
81         Point c2(x2 ,y2) ;
82         double ans = 0 ;
83         ans = Area_of_overlap(c1,R,c2,R) + Area_of_overlap(c1,r,c2,r);
84         ans -= Area_of_overlap(c1,r,c2,R) ;
85         ans -= Area_of_overlap(c1,R,c2,r) ;
86         printf("Case #%d: %.6lf
" , Case , ans) ;
87 
88     }
89     return 0;
90 }

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