hdu 5120（容斥+圆相交部分面积）

题目链接：点击打开链接

分析：求元环相交部分面积，容斥一画图就出来了；

代码如下：

#include<cmath>
#include<cstdio>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
const double esp=1e-10;

using namespace std;

struct Circle{
double x,y;
double r;
};
double calArea(Circle c1, Circle c2)
{
double d;
double s,s1,s2,s3,angle1,angle2;
d=sqrt((c1.x-c2.x)*(c1.x-c2.x)+(c1.y-c2.y)*(c1.y-c2.y));
if(d>=(c1.r+c2.r))//两圆相离
return 0;
if((c1.r-c2.r)>=d)//两圆内含,c1大
return acos(-1.0)*c2.r*c2.r;
if((c2.r-c1.r)>=d)//两圆内含,c2大
return acos(-1.0)*c1.r*c1.r;
angle1=acos((c1.r*c1.r+d*d-c2.r*c2.r)/(2*c1.r*d));
angle2=acos((c2.r*c2.r+d*d-c1.r*c1.r)/(2*c2.r*d));
s1=angle1*c1.r*c1.r;s2=angle2*c2.r*c2.r;
s3=c1.r*d*sin(angle1);
s=s1+s2-s3;
return s;
} //圆相交部分的面积
int main(){
int t;
scanf("%d",&t);
int k=1;
while(t--){
Circle a,b,c,d;
double R,r;
scanf("%lf%lf",&r,&R);
double x,y;
scanf("%lf%lf",&x,&y);
a.x=b.x=x;
a.y=b.y=y;
a.r=r;
b.r=R;
scanf("%lf%lf",&x,&y);
c.x=d.x=x;
c.y=d.y=y;
c.r=r;
d.r=R;
double ans=calArea(b,d)-calArea(b,c)-calArea(d,a)+calArea(a,c);
printf("Case #%d: %f\n",k++,ans);
}
return 0;
}