2021牛客多校2 - Girlfriend(球体积交)

Posted 2021-08-27 Frozen_Guardian

题目链接：点击查看

题目大意：空间内有6个点，满足 $|P_{1}A|\ge k_1|P_{1}B|,|P_{2}C|\ge k_2|P_{2}D|$，求 $P_1,P_2$ 各自轨迹围成的空间体的体积交

题目分析：三维坐标系中两点之商为定值的话表示的是一个球壳，叫阿波罗尼斯球壳

囤个板子，求两球体积交的

代码：

// Problem: Girlfriend
// Contest: NowCoder
// URL: https://ac.nowcoder.com/acm/contest/11253/F
// Memory Limit: 524288 MB
// Time Limit: 2000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

// #pragma GCC optimize(2)
// #pragma GCC optimize("Ofast","inline","-ffast-math")
// #pragma GCC target("avx,sse2,sse3,sse4,mmx")
#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 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);
}
int main()
{
#ifndef ONLINE_JUDGE
//	freopen("data.in.txt","r",stdin);
//	freopen("data.out.txt","w",stdout);
#endif
//	ios::sync_with_stdio(false);
	int w;
	cin>>w;
	double r1,r2,tmp1,tmp2;
	double x[10],y[10],z[10];
	double x1,x2,y1,y2,z1,z2;
	while(w--) {
		r1=0;
        r2=0;
        scanf("%lf%lf%lf",&x[2],&y[2],&z[2]);
        scanf("%lf%lf%lf",&x[1],&y[1],&z[1]);
        scanf("%lf%lf%lf",&x[4],&y[4],&z[4]);
        scanf("%lf%lf%lf",&x[3],&y[3],&z[3]);
        int k1,k2;
        scanf("%d%d",&k1,&k2);
        tmp1=(k1*x[1]+x[2])/(k1+1);
        tmp2=(k1*x[1]-x[2])/(k1-1);
        //cout << tmp1 << ' ' << tmp2 << endl;
        x1=(tmp1+tmp2)/2;
        r1+=(tmp1-tmp2)*(tmp1-tmp2);
        tmp1=(k1*y[1]+y[2])/(k1+1);
        tmp2=(k1*y[1]-y[2])/(k1-1);
        y1=(tmp1+tmp2)/2;
        r1+=(tmp1-tmp2)*(tmp1-tmp2);
        tmp1=(k1*z[1]+z[2])/(k1+1);
        tmp2=(k1*z[1]-z[2])/(k1-1);
        z1=(tmp1+tmp2)/2;
        r1+=(tmp1-tmp2)*(tmp1-tmp2);
        r1=sqrt(r1)/2;
         
        tmp1=(k2*x[3]+x[4])/(k2+1);
        tmp2=(k2*x[3]-x[4])/(k2-1);
        x2=(tmp1+tmp2)/2;
        r2+=(tmp1-tmp2)*(tmp1-tmp2);
        tmp1=(k2*y[3]+y[4])/(k2+1);
        tmp2=(k2*y[3]-y[4])/(k2-1);
        y2=(tmp1+tmp2)/2;
        r2+=(tmp1-tmp2)*(tmp1-tmp2);
        tmp1=(k2*z[3]+z[4])/(k2+1);
        tmp2=(k2*z[3]-z[4])/(k2-1);
        z2=(tmp1+tmp2)/2;
        r2+=(tmp1-tmp2)*(tmp1-tmp2);
        r2=sqrt(r2)/2;
        
        printf("%.3f\\n",sphere_intersect(x1,y1,z1,r1,x2,y2,z2,r2));
	}
	return 0;
}