題意：給定幾個點，要求覆蓋這些點的最小球半徑。(求到n個點的最大距離最小化的點
因為是單峰的所以可以用爬山演算法
主要是我的退火演算法板子精度達不到? 

//#include<bits/stdc++.h>
#include <iostream>
#include <cmath>
#include <cstdio>
#include <stdlib.h>
#include <ctime>
using namespace std;
typedef long long ll;
typedef pair<int,int> PII;
const int inf = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const int maxn = 1e4 + 5;
int n;
double X,Y;
struct point
{
    double x,y,z;
} p[maxn],pp;
double ans=1e10;
double dis(point a,point b)
{
    return ((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y)+(a.z-b.z)*(a.z-b.z));
}
int get(point x)
{
    double res=-1;
    int k;
    for(int i=1; i<=n; ++i)
    {
        double m=dis(p[i],x);
        if(m>res)
            res=m,k=i;
    }
    ans=min(ans,dis(x,p[k]));
    return k;
}
void hc()
{
    double  T=1,eps=1e-10;
    while(T>eps)
    {
        //if(pp.x<=X&&pp.y<=Y&&pp.x>=0&&pp.y>=0)
       // {
            int k=get(pp);
            pp.x=pp.x+(p[k].x-pp.x)*T;
            pp.y=pp.y+(p[k].y-pp.y)*T;
            pp.z=pp.z+(p[k].z-pp.z)*T;
       // }
        T*=0.9996;
    }
}
int  main()
{
    double X,Y;
    while(scanf("%d",&n)!=EOF)
    {
        if(n==0)break;
        pp.x=pp.y=pp.z=0;
        for(int i=1; i<=n; ++i)
         {
             cin>>p[i].x>>p[i].y>>p[i].z;
            pp.x+=p[i].x,pp.y+=p[i].y,pp.z+=p[i].z;
         }
         pp.x/=n;pp.y/=n;pp.z/=n;
        ans=1e10;
        hc();
       // printf("(%.1f,%.1f).\n",pp.x,pp.y,pp.z);
        printf("%.5f\n",sqrt(ans));
    }
   return 0;
}
 

退火沒有A掉：
 

///退火演算法
#include <iostream>
#include <cmath>
#include <algorithm>
#include <cstdio>
#include <stdlib.h>
#include <ctime>
#define mp 				make_pair
#define io 				ios::sync_with_stdio(0),cin.tie(0)
using namespace std;
typedef long long ll;
typedef pair<int, int> PII;
const int inf = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const int maxn = 1e6 + 5;
int n, x, y;
struct point { double x, y, z; } p[maxn], now, nex, ansp;
 double dis(point a, point b) { return ((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 f(point x)
{  //評估函式
	 double res = 0;
	for (int i = 1;i <= n;i++) res = max(res, dis(x, p[i]));
	return res;
}
long double ans = 1e20;//最開始的能量值,初始很大就可以,不用修改
void sa()
{
	ans = 1e22;
	 double T = 100;       //初始溫度,         (可以適當修改,最好和給的資料最大範圍相同,或者縮小其原來0.1)
	 double d = 0.9999;      //降溫係數          (可以適當修改,影響結果的精度和迴圈的次數,)
	 double eps = 1e-10;    //最終溫度          (要是因為精度問題，可以適當減小最終溫度)
	 double TT = 1.0;      //採納新解的初始概率
	 double dd = 0.999;    //(可以適當修改,採納新解變更的概率)(這個概率下面新解更新的時候,最好和未採納的新解更新的次數是一半一半)
	 double res = f(now);  //傳入的初始預設解(now)下得到的評估能量值
	if (res < ans) ans = res, ansp = now;//ansp終解
	int cnt=0;
	while (T > eps)
	{
		for (int i = -1;i <= 1;++i)
			for (int j = -1;j <= 1;++j)
				if ((now.x+T*i<=100)&&(now.x+T*i>=0)&&(now.y+T*j<=100)&&(now.y+T*j>=0))
				{
					nex.x = now.x + T * i, nex.y = now.y + T * j;//新解
					 double tmp = f(nex);//新解下的評估能量值
					if (tmp < ans) ans = tmp, ansp = nex;//降溫成功,更新當前最優解

					if (tmp < res) res = tmp, now = nex;// 降溫成功,採納新解
					else if (TT > rand() % 10000 / 10000.0) res = tmp, now = nex;//,cout<<"======"<<endl;//沒有 降溫成功，但是以一定的概率採納新解
					//else cout<<"="<<endl;//用於測試，設定的採納新解的概率,是否為一半一半,可以適當修改降溫引數dd
				}cnt++;
		T *= d; TT *= dd;
	}
	//cout<<cnt<<endl;
}
int main()
{
	srand(time(0));
	while (scanf("%d", &n)!=EOF) {
		if (n == 0)break;
		now.x = now.y = 0, now.z = 0;
		for (int i = 1;i <= n;++i)
		{
			cin >> p[i].x >> p[i].y >> p[i].z;//scanf("%lf%lf%lf",&p[i].x,&p[i].y),
			now.x += p[i].x, now.y += p[i].y, now.z += p[i].z;
		}now.x = now.y = 55, now.z = 55;sa();
		//now.x /= n, now.y /= n, now.z /= n;sa();

		  //now.x = now.y = 20, now.z = 20;sa();
		printf("%.5Lf\n", sqrt(ans));//cout<<ans<<endl;
	}
	return 0;
}