HDU5120 （容斥原理）

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题目大意：求两环的相交面积。

容斥原理，两个大圆面积之交集 - 两个大圆与另一个小圆面积之交集 + 两小圆面积之交集。直接拿模板。

#include<iostream>#include<math.h>#include<iomanip>using namespace std;const double pai=acos(-1.0);class Circle{public:    double x,y,r;    Circle(double x1,double Y1,double R):x(x1),y(Y1),r(R){}    double intersection(Circle b)    {        double dis=sqrt( (x-b.x)*(x-b.x)+(y-b.y)*(y-b.y) );        if(dis>=r+b.r)            return 0;        double minr=r<b.r?r:b.r;        double maxr=r>b.r?r:b.r;        if(dis<=maxr-minr)            return pai*minr*minr;        double ang1=acos( (dis*dis+r*r-b.r*b.r)/(2*dis*r) );        double ang2=acos( (dis*dis+b.r*b.r-r*r)/(2*dis*b.r) );        double ans=0;        ans-=dis*r*sin(ang1)*0.5*2;        ans+=ang1/(pai*2)*pai*r*r*2+ang2/(pai*2)*pai*b.r*b.r*2;        return ans;    }};int main(){    int times;    cin>>times;    for(int time=1;time<=times;++time)    {         double r1,r2,x1,x2,y1,y2;         cin>>r1>>r2>>x1>>y1>>x2>>y2;         Circle a(x1,y1,r2);         Circle A(x1,y1,r1);         Circle B(x2,y2,r1);         Circle b(x2,y2,r2);        cout<<"Case #"<<time<<": ";        cout<<fixed<<setprecision(6)<< A.intersection(B)-A.intersection(b)-B.intersection(a)+a.intersection(b)<<endl;    }}

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