题目链接：https://nanti.jisuanke.com/t/33673

思路：就是一个最小球覆盖，模拟退火，和poj2069要求的一样，还可以三分套三分或者几何法。。。

#pragma GCC optimize(2)
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <bitset>
#include <cmath>
#include <cctype>
#include <iostream>
#include <algorithm>
#include <string>
#include <vector>
#include <queue>
#include <map>
#include <set>
#include <sstream>
#include <iomanip>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const ll inff = 0x3f3f3f3f3f3f3f3f;
#define FOR(i,a,b) for(int i(a);i<=(b);++i)
#define FOL(i,a,b) for(int i(a);i>=(b);--i)
#define REW(a,b) memset(a,b,sizeof(a))
#define inf int(0x3f3f3f3f)
#define si(a) scanf("%d",&a)
#define sl(a) scanf("%I64d",&a)
#define sd(a) scanf("%lf",&a)
#define ss(a) scanf("%s",a)
#define mod ll(1e9+7)
#define pb push_back
#define eps 1e-6
#define lc d<<1
#define rc d<<1|1
#define Pll pair<ll,ll>
#define P pair<int,int>
#define pi acos(-1)
struct point{
double x,y,z;}a[108];
int n;
double dis(point a,point b)
{
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y)+(a.z-b.z)*(a.z-b.z));
}
double slove()
{
    double temp=10000,ans=inf,ma;
    point qw;
    qw.x=qw.y=qw.z=0;
    int s=1;
    while(temp>eps)
    {
        FOR(i,1,n)
         if(dis(qw,a[s])<dis(qw,a[i])) s=i;
        ma=dis(qw,a[s]);
        ans=min(ans,ma);
        qw.x+=(a[s].x-qw.x)/ma*temp;
        qw.y+=(a[s].y-qw.y)/ma*temp;
        qw.z+=(a[s].z-qw.z)/ma*temp;
        temp*=0.98;
    }
    return ans;
}
int main()
{
    cin.tie(0);
    cout.tie(0);
    cin>>n;
    FOR(i,1,n) sd(a[i].x),sd(a[i].y),sd(a[i].z);
    printf("%.5f\n",slove());
    return 0;
}