[球体积交]Girlfriend 2021牛客多校第2场 F

题目描述

ZYT and ZFeT are great PUAs and both have two girlfriends,but they never think that their girlfriends could be so great at magic.

One day, when ZYT and ZFeT were busy training PU skills,they suddenly found themselves spelled by their girlfriends.

And their girlfriends told them that they would be freed after 1010010100 years .

After examining carefully , they found the mechanism of the spell:

ZYT must keep the distance between himself and his lover A no less than k1k1 times the distance between himself and his lover B.
ZFeT must keep the distance between himself and his lover C no less than k2k2 times the distance between himself and his lover D.

Let P1,P2P1,P2 denote the positions of ZYT and ZFeT, so that is:

∣AP1∣≥k1∣BP1∣,∣CP2∣≥k2∣DP2∣∣AP1∣≥k1∣BP1∣,∣CP2∣≥k2∣DP2∣

As they immediately discovered:their available living places formed two geometric forms in space and they may has intersection inside.
Surely the intersection was far too enough for them to become more and more gay(I mean,happy) in these 1010010100 years.
Wondered how gay they could be,they wanted to ask you the volume of of the intersection.

输入描述:

The first line of input contains a single integer TT — the number of test cases.Each test case consists of five lines:The first four lines contains three integers (xi,yi,zi)(xi,yi,zi), denoting the coordinates of points A,B,C,D.Then another line contains two integers k1,k2k1,k2, described as above.

输出描述:

Output TT lines denoting the answer of each test case, answer with an error less than 10−310−3 (relatively or absolutely) will be accepted.

示例1

输入

输出

0.262

题意： 给出三维空间四个点A、B、C、D的坐标，以及k1和k2，第一个人的活动范围是所有到A距离大于等于k1倍的到B距离的位置，第二个人的活动范围是所有到C距离大于等于k2倍的到D距离的位置，求二者活动范围交集的体积。

分析： 列出方程后可以发现每个人的活动范围都是一个球体，所以问题就转化为求两球体的体积交了，主要还是公式化简比较麻烦，当求出球心坐标和半径后就可以套用模板了。

具体代码如下：

#include<iostream>
#include<cstdio>
#include<string>
#include<ctime>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<stack>
#include<climits>
#include<queue>
#include<map>
#include<set>
#include<sstream>
#include<cassert>
#include<bitset>
#include<list>
#include<unordered_map>
#define double long double
#define lowbit(x) (x&-x)
using namespace std;
typedef long long LL;
typedef unsigned long long ull;
template<typename T>
inline void read(T &x)
{T f=1;x=0;char ch=getchar();while(0==isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}while(0!=isdigit(ch)) x=(x<<1)+(x<<3)+ch-'0',ch=getchar();x*=f;
}
template<typename T>
inline void write(T x)
{if(x<0){x=~(x-1);putchar('-');}if(x>9)write(x/10);putchar(x%10+'0');
}
const int inf=0x3f3f3f3f;
const int N=1e6+100;
const double pi=acos(-1);double pow2(double x){return x*x;}double pow3(double x){return x*x*x;}double dis(double x1,double y1,double z1,double x2,double y2,double z2)
{return pow2(x1-x2)+pow2(y1-y2)+pow2(z1-z2);
}double cos(double a,double b,double c){return (b*b+c*c-a*a)/(2*b*c);}double cap(double r,double h){return pi*(r*3-h)*h*h/3;}//2球体积交
double sphere_intersect(double x1,double y1,double z1,double r1,double x2,double y2,double z2,double r2)
{double d=dis(x1,y1,z1,x2,y2,z2);//相离if(d>=pow2(r1+r2))return 0;//包含if(d<=pow2(r1-r2))return pow3(min(r1,r2))*4*pi/3;//相交double h1=r1-r1*cos(r2,r1,sqrt(d)),h2=r2-r2*cos(r1,r2,sqrt(d));return cap(r1,h1)+cap(r2,h2);
}//2球体积并
double sphere_union(double x1,double y1,double z1,double r1,double x2,double y2,double z2,double r2)
{double d=dis(x1,y1,z1,x2,y2,z2);//相离if(d>=pow2(r1+r2))return (pow3(r1)+pow3(r2))*4*pi/3;//包含if(d<=pow2(r1-r2))return pow3(max(r1,r2))*4*pi/3;//相交double h1=r1+r1*cos(r2,r1,sqrt(d)),h2=r2+r2*cos(r1,r2,sqrt(d));return cap(r1,h1)+cap(r2,h2);
}void solve()
{double x[5],y[5],z[5];double x1, y1, z1, r1, x2, y2, z2, r2;for(int i=1;i<=4;i++) cin>>x[i]>>y[i]>>z[i];double k1,k2;cin>>k1>>k2;x1 = ((k1*k1*x[2]-x[1])/(k1*k1-1));y1 = ((k1*k1*y[2]-y[1])/(k1*k1-1));z1 = ((k1*k1*z[2]-z[1])/(k1*k1-1));r1 = (x1*x1+y1*y1+z1*z1-(k1*k1*(x[2]*x[2]+y[2]*y[2]+z[2]*z[2]))/(k1*k1-1)+(x[1]*x[1]+y[1]*y[1]+z[1]*z[1])/(k1*k1-1));r1 = sqrt(r1);k1 = k2;x[1] = x[3];y[1] = y[3];z[1] = z[3];x[2] = x[4];z[2] = z[4];y[2] = y[4];x2 = ((k1*k1*x[2]-x[1])/(k1*k1-1));y2 = ((k1*k1*y[2]-y[1])/(k1*k1-1));z2 = ((k1*k1*z[2]-z[1])/(k1*k1-1));r2 = (x2*x2+y2*y2+z2*z2-(k1*k1*(x[2]*x[2]+y[2]*y[2]+z[2]*z[2])/(k1*k1-1))+(x[1]*x[1]+y[1]*y[1]+z[1]*z[1])/(k1*k1-1));r2 = sqrt(r2);
//  printf("%Lf %Lf %Lf %Lf\n", x1, y1, z1, r1);
//  printf("%Lf %Lf %Lf %Lf\n", x2, y2, z2, r2);double v = sphere_intersect(x1, y1, z1, r1, x2, y2, z2, r2);printf("%Lf\n", v);
}
int main(){int _;scanf("%d",&_);while(_--) solve();return 0;
}/*
10
0 0 0
1 1 1
1 0 1
1 1 0
2 3
*/