经典圆交面积求解之“Intersection ”

题目大意：给你两个相同的圆环，求出它们相交的面积。输入数据为圆环内径，外径以及两个圆环的圆心坐标（以上数据均为整数）。

解题思路：两圆环相交面积 = 两个大圆相交面积 - 两倍大小圆相交面积 + 两个小圆相交面积

AC代码：

 1 #include <cstdlib>
 2 #include <iostream>
 3 #include <cmath>
 4 #include <cstring>
 5 #include <algorithm>
 6 #include <stdio.h>
 7 using namespace std;
 8 const double PI = acos(-1);
 9 
10 struct circle {
11        double x;
12        double y;
13        double r;
14 };
15 double dist(circle a, circle b)
16 {
17     return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
18 }
19 double area(circle a, circle b) {
20 
21        if((dist(a, b)+min(a.r,b.r))<=max(a.r,b.r)) 
22         {
23             if(a.r<b.r)
24                  return PI*a.r*a.r;
25             else
26                  return PI*b.r*b.r;
27         }
28         else if(dist(a, b)>=(a.r+b.r)) 
29             return 0.0;
30         else
31         {
32             double length=dist(a, b);
33             double d1=2*acos((a.r*a.r+length*length-b.r*b.r)/(2*a.r*length));
34             double d2=2*acos((b.r*b.r+length*length-a.r*a.r)/(2*b.r*length));
35             double area1=a.r*a.r*d1/2-a.r*a.r*sin(d1)/2;
36             double area2=b.r*b.r*d2/2-b.r*b.r*sin(d2)/2;
37             double area=area1+area2;
38             return area;
39         }
40 }
41 int main()
42 {
43     int T;
44     circle a, b, A, B;
45     double r, R;
46     scanf("%d",&T);
47     for(int i=1; i<=T; i++)
48     {
49         scanf("%lf%lf", &r, &R);
50         a.r = b.r = r;
51         A.r = B.r = R;
52         scanf("%lf%lf", &a.x, &a.y);
53         A.x = a.x;
54         A.y = a.y;
55         scanf("%lf%lf", &b.x, &b.y);
56         B.x = b.x;
57         B.y = b.y;
58         double sec = area(A, B) - area(A, b) - area(B, a) + area(a, b);
59         printf("Case #%d: %.6lf
", i, sec);
60 
61     }
62 
63     return 0;
64 }

不知为何用Java代码会超时；但是思路还是一样的。