1. 程式人生 > >hdu 5120 Intersection(求相交圓的面積)

hdu 5120 Intersection(求相交圓的面積)

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;
 double S1,S2,S3,a1,a2,d,p,ans,mi;
 int x1,yy,x2,y2;
struct circle
{
    double x,y,r;
};

double area(int rrr,int RRR)//兩圓相交的面積
{
      circle a,b;
      a.x=x1;
      a.y=yy;
      b.x=x2;
      b.y=y2;
      a.r=rrr;
      b.r=RRR;
    d=(double)sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
        mi=min(a.r,b.r);
        if(d>a.r+b.r||d==a.r+b.r)
            ans=0.0;
        else if(d<abs(a.r-b.r)||d==abs(a.r-b.r))
            ans=acos(-1.0)*mi*mi;
        else
        {
            p=(a.r+b.r+d)/2.0;
            a1=acos((a.r*a.r+d*d-b.r*b.r)/(2.0*a.r*d));
            a2=acos((b.r*b.r+d*d-a.r*a.r)/(2.0*b.r*d));
            S1=a1*a.r*a.r;
            S2=a2*b.r*b.r;
            S3=2*sqrt(p*(p-a.r)*(p-b.r)*(p-d));
            ans=S1+S2-S3;
        }
        return ans;
}

int main()
{

    int casen;
    cin>>casen;
    for(int tt=1;tt<=casen;tt++)
    {
        int rr,RR;
        cin>>rr>>RR;
        cin>>x1>>yy>>x2>>y2;
        double BB=area(RR,RR);
        double Bb=area(RR,rr);
        double bB=area(rr,RR);
        double bb=area(rr,rr);
        double Ans=BB-Bb-bB+bb;
        printf("Case #%d: %.6lf\n",tt,Ans);
    }
    return 0;
}