hdu 3756 Dome of Circus(三分)

题意：求能将所有点覆盖的最小体积的圆锥。

体积变化为凹形，故可以三分半径r，再枚举各个点找最大的高h。

/*
Author:Zhaofa Fang
Lang:C++
*/
#include <cstdio>
#include <cstdlib>
#include <sstream>
#include <iostream>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <string>
#include <utility>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
using namespace std;

typedef long long ll;
#define DEBUG(x) cout<< #x << ':' << x << endl
#define REP(i,n) for(int i=0;i < (n);i++)
#define REPD(i,n) for(int i=(n-1);i >= 0;i--)
#define FOR(i,s,t) for(int i = (s);i <= (t);i++)
#define FORD(i,s,t) for(int i = (s);i >= (t);i--)
#define PII pair<int,int>
#define PB push_back
#define MP make_pair
#define ft first
#define sd second
#define lowbit(x) (x&(-x))
#define INF (1<<30)


const double PI = 3.1415926;
int n;
struct point
{
    double x,y,z;
}P[10005];
double volume(double r,double h)
{
    return PI*r*r*h/3.0;
}

double fun(double r,double &hx)
{
    double MAX = 0;
    REP(i,n)
    {
        double h = P[i].z*r/(r-sqrt(P[i].x*P[i].x+P[i].y*P[i].y));
        MAX = max(MAX,h);
    }
    hx = MAX;
    return volume(r,MAX);
}
int main()
{
    //freopen("in","r",stdin);
    //freopen("out","w",stdout);
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        double M=0;
        REP(i,n)
        {
            scanf("%lf%lf%lf",&P[i].x,&P[i].y,&P[i].z);
            M = max(M,sqrt(P[i].x*P[i].x+P[i].y*P[i].y));
        }
        double l=M+1e-6,r=100000.0,hx;
        while(fabs(l-r) > 1e-6)
        {
            double lm = l + (r-l)/3.0;
            double rm = r - (r-l)/3.0;
            if(fun(lm,hx)<fun(rm,hx))r = rm;
            else l = lm;
        }
        printf("%.3f %.3f\n",hx,r);
    }
    return 0;
}