hdu 5120（容斥+圆相交部分面积）

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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;}

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