题目链接2021牛客暑期多校训练营2 F Girlfriend

数学题
意思就是求两个球相交部分的体积
上代码

#include <bits/stdc++.h>
using namespace std;

const double PI = acos(-1);

//坐标
struct pos
{
    double x,y,z;
};
//输入数据
int T;
pos p[5];
double k1,k2;

//由两球球心和半径得到其相交部分体积
void calc(pos o1,pos o2,double r1,double r2)
{
    double ans=0.0;
    double dis=sqrt((o1.x-o2.x)*(o1.x-o2.x)+(o1.y-o2.y)*(o1.y-o2.y)+(o1.z-o2.z)*(o1.z-o2.z));

    //不相交
    if(dis>=r1+r2)
        ans=0;
    //内含
    else if (dis+r1<=r2)
        ans=(4.00/3.00)*PI*r1*r1*r1;
    else if(dis+r2<=r1)
        ans=(4.00/3.00)*PI*r2*r2*r2;
    else
    {
    	//余弦定理
        double cos_r1_dis=(r1*r1+dis*dis-r2*r2)/(2.0*dis*r1);
		//计算球缺的高度
        double h1=r1-r1*cos_r1_dis;
		//球缺体积计算公式
        ans+=PI*h1*h1*(r1-h1/3.0);
        
		//余弦定理
        double cos_r2_dis=(r2*r2+dis*dis-r1*r1)/(2.0*dis*r2);
		//计算球缺的高度
        double h2=r2-r2*cos_r2_dis;
		//球缺体积计算公式
        ans+=(PI*h2*h2*(r2-h2/3.0));
    }
    printf("%.3f\n",ans);
}

int main()
{
    cin>>T;
    while(T--)
    {
        for(int i = 0; i < 4; i++)
            cin >> p[i].x >> p[i].y >> p[i].z;

        cin>>k1>>k2;
        double tmp1=k1*k1-1,tmp2=k2*k2-1;
		
        pos O1,O2;
        double r1,r2,D1,D2;

        //球1
        O1.x=(k1*k1*p[1].x-p[0].x)/tmp1;
        O1.y=(k1*k1*p[1].y-p[0].y)/tmp1;
        O1.z=(k1*k1*p[1].z-p[0].z)/tmp1;


        D1=k1*k1*(p[1].x*p[1].x+p[1].y*p[1].y+p[1].z*p[1].z)-p[0].x*p[0].x-p[0].y*p[0].y-p[0].z*p[0].z;
        D1/=tmp1;

        r1=sqrt(O1.x*O1.x+O1.y*O1.y+O1.z*O1.z-D1);

        //球2
        O2.x=(k2*k2*p[3].x-p[2].x)/tmp2;
        O2.y=(k2*k2*p[3].y-p[2].y)/tmp2;
        O2.z=(k2*k2*p[3].z-p[2].z)/tmp2;


        D2=k2*k2*(p[3].x*p[3].x+p[3].y*p[3].y+p[3].z*p[3].z)-p[2].x*p[2].x-p[2].y*p[2].y-p[2].z*p[2].z;
        D2/=tmp2;

        r2=sqrt(O2.x*O2.x+O2.y*O2.y+O2.z*O2.z-D2);
        //printf("o1: %lf %lf %lf r1= %lf\n",O1.x,O1.y,O1.z,r1);
        //printf("o2: %lf %lf %lf r2= %lf\n",O2.x,O2.y,O2.z,r2);
        calc(O1,O2,r1,r2);
    }
    return 0;
}