poj、 2546 两圆面积交

验一下模板。。。。

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
const double eps = 1e-8;
const double pi = acos(-1.0);
struct Point {
    double x, y;
    Point operator - (const Point& t) const {
        Point tmp;
        tmp.x = x - t.x;
        tmp.y = y - t.y;
        return tmp;
    }
    Point operator + (const Point& t) const {
        Point tmp;
        tmp.x = x + t.x;
        tmp.y = y + t.y;
        return tmp;
    }
    bool operator == (const Point& t) const {
        return fabs(x-t.x) < eps && fabs(y-t.y) < eps;
    }
}GP; 

double cir_area_inst(Point c1, double r1, Point c2, double r2) {			// 两圆面积交
	double a1, a2, d, ret;
    d = sqrt((c1.x-c2.x)*(c1.x-c2.x)+(c1.y-c2.y)*(c1.y-c2.y));
    if ( d > r1 + r2 - eps ) 
        return 0;
    if ( d < r2 - r1 + eps ) 
        return pi*r1*r1;
	if ( d < r1 - r2 + eps )
		return pi*r2*r2;
    a1 = acos((r1*r1+d*d-r2*r2)/2/r1/d);
    a2 = acos((r2*r2+d*d-r1*r1)/2/r2/d);
    ret = (a1-0.5*sin(2*a1))*r1*r1 + (a2-0.5*sin(2*a2))*r2*r2;
	return ret;
}
int main()
{
    Point a,b;
	double r1,r2;
	scanf("%lf%lf%lf%lf%lf%lf",&a.x,&a.y,&r1,&b.x,&b.y,&r2);
    double area = cir_area_inst(a,r1,b,r2);
	printf("%.3lf\n",area);
	return 0;
}