HDOJ 5120 Intersection（求两圆相交面积）

#include <iostream>
#include <stdlib.h>
#include <stdio.h>
#include <math.h>
using namespace std;
#define eps 1e-8
#define pi acos(-1.0)

int sig(double x)
{
return (x > eps) - (x < -eps);
}

typedef struct Point
{
double x, y;
Point() {}
Point(double _x, double _y):
x(_x), y(_y) {}
Point operator -(const Point &argu) const
{
return Point(x - argu.x, y - argu.y);
}
bool operator <(const Point &argu) const
{
if(sig(x - argu.x) == 0)
return y < argu.y;
return x < argu.x;
}
double dis(const Point &argu) const
{
return sqrt((x - argu.x) * (x - argu.x) + (y - argu.y) * (y - argu.y));
}
}pp;

typedef struct Circle
{
Point o;
double r;
Circle() {}
Circle(Point _o, double _r):
o(_o), r(_r) {}
bool operator <(const Circle &argu) const
{
if(sig(r - argu.r) == 0)
return o < argu.o;
else
return r < argu.r;

}
///参数圆半径更小些
double Intersection_Area(const Circle &argu) const
{
double d = ((*this).o).dis(argu.o);

if(sig(d - r - argu.r) >= 0)
return 0;
if(sig(r - argu.r - d) >= 0 || sig(d) == 0)
return pi * argu.r * argu.r;
double ang1 = acos((argu.r * argu.r + d * d - r * r) / (2 * argu.r * d));
double ang2 = acos((r * r + d * d - argu.r * argu.r) / (2 * r * d));
return ang1 * argu.r * argu.r + ang2 * r * r - argu.r * d * sin(ang1);
}
}cc;

cc c1[2], c2[2];

double calc()
{
scanf("%lf %lf", &c1[0].r, &c1[1].r);
c2[0].r = c1[0].r, c2[1].r = c1[1].r;

scanf("%lf%lf", &c1[0].o.x, &c1[0].o.y);
c1[1].o.x = c1[0].o.x, c1[1].o.y = c1[0].o.y;

scanf("%lf%lf", &c2[0].o.x, &c2[0].o.y);
c2[1].o.x = c2[0].o.x, c2[1].o.y = c2[0].o.y;

return c1[1].Intersection_Area(c2[1]) - c1[1].Intersection_Area(c2[0]) - c2[1].Intersection_Area(c1[0]) + c1[0].Intersection_Area(c2[0]);
}

int main()
{
//    freopen("5120.in", "r", stdin);

int t;
scanf("%d", &t);
for(int cas = 1; cas <= t; cas++)
{
printf("Case #%d: %.6lf\n", cas, calc());
}
return 0;
}