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：

#include <cstdlib>
#include <iostream>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <stdio.h>
using namespace std;
const double PI = acos(-1);

struct circle {
       double x;
       double y;
       double r;
};
double dist(circle a, circle b)
{
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
double area(circle a, circle b) {

       if((dist(a, b)+min(a.r,b.r))<=max(a.r,b.r)) 
        {
            if(a.r<b.r)
                 return PI*a.r*a.r;
            else
                 return PI*b.r*b.r;
        }
        else if(dist(a, b)>=(a.r+b.r)) 
            return 0.0;
        else
        {
            double length=dist(a, b);
            double d1=2*acos((a.r*a.r+length*length-b.r*b.r)/(2*a.r*length));
            double d2=2*acos((b.r*b.r+length*length-a.r*a.r)/(2*b.r*length));
            double area1=a.r*a.r*d1/2-a.r*a.r*sin(d1)/2;
            double area2=b.r*b.r*d2/2-b.r*b.r*sin(d2)/2;
            double area=area1+area2;
            return area;
        }
}
int main()
{
    int T;
    circle a, b, A, B;
    double r, R;
    scanf("%d",&T);
    for(int i=1; i<=T; i++)
    {
        scanf("%lf%lf", &r, &R);
        a.r = b.r = r;
        A.r = B.r = R;
        scanf("%lf%lf", &a.x, &a.y);
        A.x = a.x;
        A.y = a.y;
        scanf("%lf%lf", &b.x, &b.y);
        B.x = b.x;
        B.y = b.y;
        double sec = area(A, B) - area(A, b) - area(B, a) + area(a, b);
        printf("Case #%d: %.6lf
", i, sec);

    }

    return 0;
}

不知为何用Java代码会超时；但是思路还是一样的。