hdu 5120 Intersection 2014ACM/ICPC亚洲区北京站-重现赛

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画个图就可以用容斥原理推出公式。

area=circle1大圆和circle2大圆面积交-circle1大圆和circle2小圆面积交-circle1小圆和circle2大圆面积交+circle1小圆和circle2小圆面积交

然后看一下圆心重合、相切、分离的几种case能否满足即可，这里的模板比之前的多加了个两个圆重合的特判。

另外在求圆心距的时候，要注意圆心重合时要特判return 0，如果用sqrt会返回nan（虽然开始没理这个也A了==）。

#include<iostream>#include<stdio.h>#include<cstdio>#include<stdlib.h>#include<vector>#include<string>#include<cstring>#include<cmath>#include<algorithm>#include<stack>#include<queue>#include<ctype.h>#include<map>#include<time.h>#include<bitset>using namespace std;//hdu 5120int T;double r;double R;double x[2];double y[2];double ans;const double eps=1e-9;class circle{public:    double x;    double y;    double r;    circle(){x=0;y=0;r=0;}    circle(double xx,double yy,double rr)    {        x=xx;        y=yy;        r=rr;    }};circle cir[25];double pi=acos(-1.0);double dis(circle r1,circle r2){    double xd=r1.x-r2.x;    double yd=r1.y-r2.y;    //cout<<"dis "<<xd<<" "<<yd<<endl;    if(abs(xd)<eps&&abs(yd)<eps) return 0;    else return sqrt(xd*xd+yd*yd);}double intersection(circle c1,circle c2){    double d=dis(c1,c2);    //cout<<"d: "<<d<<endl;    if(d<eps) return min(pi*c1.r*c1.r,pi*c2.r*c2.r);//特判两个圆重合的case    else if(d>=(c1.r+c2.r)) return 0;    else if(d<fabs(c1.r-c2.r))    {       // cout<<"haha  "<<d<<" "<<fabs(c1.r-c2.r)<<endl;        //cout<<"area: "<<pi*c1.r*c1.r<<" "<<pi*c2.r*c2.r<<endl;        return min(pi*c1.r*c1.r,pi*c2.r*c2.r);    }    else    {        double h=(c1.r+c2.r+d)/2.0;        double s=sqrt(h*(h-c1.r)*(h-c2.r)*(h-d))*2.0;//三角形面积的海伦公式        double angle1=acos((c1.r*c1.r+d*d-c2.r*c2.r)/2.0/c1.r/d);//余弦定理        double angle2=acos((c2.r*c2.r+d*d-c1.r*c1.r)/2.0/c2.r/d);        double s1=angle1*c1.r*c1.r;        double s2=angle2*c2.r*c2.r;//求扇形面积        return s1+s2-s;//inclusive and declusive principle    }}int main(){    freopen("input.txt","r",stdin);    //freopen("data.txt","r",stdin);    //freopen("out1.txt","w",stdout);    scanf("%d",&T);    for(int ca=1;ca<=T;ca++)    {        scanf("%lf %lf",&r,&R);        scanf("%lf %lf",&x[0],&y[0]);        scanf("%lf %lf",&x[1],&y[1]);        ans=0;        circle c1=circle(x[0],y[0],r);        circle c2=circle(x[0],y[0],R);        circle c3=circle(x[1],y[1],r);        circle c4=circle(x[1],y[1],R);        //cout<<"inter: "<<intersection(c2,c4)<<" "<<intersection(c1,c4)<<" "<<intersection(c2,c3)<<" "<<intersection(c1,c3)<<endl;        ans=intersection(c2,c4)-intersection(c1,c4)-intersection(c2,c3)+intersection(c1,c3);        printf("Case #%d: %.6lf\n",ca,ans);    }     return 0;}

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