【 2018南京】Country Meow （模拟退火）

In the $24th24^{\text{th}}24th century, there is a country somewhere in the universe, namely Country Meow. Due to advanced technology, people can easily travel in the 3-dimensional space.$

There are $(xi,yi,zi)(x_i,y_i,z_i)(xi,yi,zi) in Cartesian coordinate.$

Due to the increasing threat from Country Woof, the president decided to build a new combatant command, so that troops in different cities can easily communicate. Hence, the Euclidean distance between the combatant command and any city should be minimized.

Your task is to calculate the minimum Euclidean distance between the combatant command and the farthest city.

Input

The first line contains an integer $\le N \le 1001≤N≤100).$

The following $xi,yi,zix_i, y_i, z_ixi,yi,zi (−100000≤xi,yi,zi≤100000-100000 \le x_i, y_i, z_i \le 100000−100000≤xi,yi,zi≤100000).$

Output

Print a real number $—\text{---}— the minimum Euclidean distance between the combatant command and the farthest city. Your answer is considered correct if its absolute or relative error does not exceed 10−310^{-3}10−3. Formally, let your answer be aaa, and the jury's answer be bbb. Your answer is considered correct if ∣a−b∣max⁡(1,∣b∣)≤10−3\frac{|a - b|}{\max(1, |b|)} \le 10^{-3}max(1,∣b∣)∣a−b∣≤10−3.$

本题答案不唯一，符合要求的答案均正确

样例输入1

样例输出1

2.500000590252103

样例输入2

样例输出2

0.816496631812619


SOLUTION:

初始化0，0，然后每次向最远的点移动一定的距离
感觉像是退火，但又不是，

CODE:

#include<bits/stdc++.h>
#define LL long long
using namespace std;
const int INF=0x3f3f3f3f;
const int maxn=150;
const double eps=1e-3;       //精度
const double start_T=1000;   //初始温度
const double rate=0.98;      //温度下降速率
struct point
{
    double x;
    double y;
    double z;
} p[maxn];
int N;
double dist(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 solve()
{
    double T=start_T;
    point ans_p={0,0,0};  //初始点
    double ans=1e99;      //预设一个较大值
    while(T>eps)
    {
        point maxd_p=p[1];
        for(int i=2;i<=N;i++)
        {
            if(dist(ans_p,p[i])>dist(ans_p,maxd_p))
                maxd_p=p[i];
        }
        //找到距离ans_p最远的点,maxd_p
        ans=min(ans,dist(ans_p,maxd_p));
        ans_p.x+=(maxd_p.x-ans_p.x)*(T/start_T);    //以一定概率靠近maxd_p
        ans_p.y+=(maxd_p.y-ans_p.y)*(T/start_T);
        ans_p.z+=(maxd_p.z-ans_p.z)*(T/start_T);
        T*=rate;
    }
    return ans;
}
int main()
{
    scanf("%d",&N);
    for(int i=1;i<=N;i++)
        scanf("%lf %lf %lf",&p[i].x,&p[i].y,&p[i].z);
    printf("%.8lf",solve());
    return 0;
}