zoj 1597 Circular Area (求两园相交面积)

# include <algorithm>
# include <stdio.h>
# include <string.h>
# include <iostream>
# include <math.h>
using namespace std;
# define PI acos(-1.0)
const double eps = 1e-8;
struct  node
{
double x;
double y;
double r;
};
struct node a1[10];

double dis(double x1,double y1,double x2,double y2)
{
return sqrt( (x1-x2)*(x1-x2)+(y1-y2)*(y1-y2) );
}
double area()///不相离不相含求相交面积
{
double a = dis(a1[1].x,a1[1].y,a1[0].x,a1[0].y), b = a1[0].r, c = a1[1].r;
double cta1 = acos((a * a + b * b - c * c) / 2 / (a * b)),
cta2 = acos((a * a + c * c - b * b) / 2 / (a * c));
double s1 = a1[0].r*a1[0].r*cta1 - a1[0].r*a1[0].r*sin(cta1)*(a * a + b * b - c * c) / 2 / (a * b);
double s2 = a1[1].r*a1[1].r*cta2 - a1[1].r*a1[1].r*sin(cta2)*(a * a + c * c - b * b) / 2 / (a * c);
return s1 + s2;
}
int main()
{
double x1,x2,y1,y2,xx1,xx2,yy1,yy2;
double d1,d2,d3,d4,d5,d6,d7,d8;
while(~scanf("%lf %lf %lf %lf %lf %lf",&a1[0].x,&a1[0].y,&a1[0].r,&a1[1].x,&a1[1].y,&a1[1].r))
{
if(dis(a1[0].x,a1[0].y,a1[1].x,a1[1].y)<=fabs(a1[0].r-a1[1].r))///相包含
{
if(a1[1].r>a1[0].r)
printf("%.3lf\n",PI*a1[0].r*a1[0].r);
else
printf("%.3lf\n",PI*a1[1].r*a1[1].r);
}
else if(dis(a1[0].x,a1[0].y,a1[1].x,a1[1].y)>=a1[0].r+a1[1].r)///相离
printf("0.000\n");
else ///相交
printf("%.3lf\n",area());
}
return 0;
}