POJ - 2069 Super Star【爬山算法 || 模拟退火】

Time limit 1000 ms
Memory limit 65536 kB

During a voyage of the starship Hakodate-maru (see Problem 1406), researchers found strange synchronized movements of stars. Having heard these observations, Dr. Extreme proposed a theory of “super stars”. Do not take this term as a description of actors or singers. It is a revolutionary theory in astronomy.
According to this theory, starts we are observing are not independent objects, but only small portions of larger objects called super stars. A super star is filled with invisible (or transparent) material, and only a number of points inside or on its surface shine. These points are observed as stars by us.

In order to verify this theory, Dr. Extreme wants to build motion equations of super stars and to compare the solutions of these equations with observed movements of stars. As the first step, he assumes that a super star is sphere-shaped, and has the smallest possible radius such that the sphere contains all given stars in or on it. This assumption makes it possible to estimate the volume of a super star, and thus its mass (the density of the invisible material is known).

You are asked to help Dr. Extreme by writing a program which, given the locations of a number of stars, finds the smallest sphere containing all of them in or on it. In this computation, you should ignore the sizes of stars. In other words, a star should be regarded as a point. You may assume the universe is a Euclidean space.

Input

The input consists of multiple data sets. Each data set is given in the following format.

n
x1 y1 z1
x2 y2 z2
…
xn yn zn

The first line of a data set contains an integer n, which is the number of points. It satisfies the condition 4 <= n <= 30.

The location of n points are given by three-dimensional orthogonal coordinates: (xi, yi, zi) (i = 1, …, n). Three coordinates of a point appear in a line, separated by a space character. Each value is given by a decimal fraction, and is between 0.0 and 100.0 (both ends inclusive). Points are at least 0.01 distant from each other.

The end of the input is indicated by a line containing a zero.

Output

For each data set, the radius of the smallest sphere containing all given points should be printed, each in a separate line. The printed values should have 5 digits after the decimal point. They may not have an error greater than 0.00001.

题目大意

在三维坐标系中给出n个点，用一个球体包含这n个点，求最小半径

题目分析

爬山算法 x 模拟退火

假设当前温度为T
在n个点中找到与当前记录解 $(x,y,z)$ 最远的点 $(x_0,y_0,z_0)$
假设他们距离为 $dis$
那么另 $(x,y,z)$ 向其移动

x+=(x0-x)/dis*T;
y+=(y0-y)/dis*T;
z+=(z0-z)/dis*T;

由于温度不断减小
每次移动的范围也逐渐减小

#include<iostream>
#include<cmath>
#include<algorithm>
#include<queue>
#include<cstring>
#include<cstdio>
using namespace std;
typedef double dd;
#define T 100
#define delta 0.98
#define eps 1e-8

int read()
{
    int f=1,x=0;
    char ss=getchar();
    while(ss<'0'||ss>'9'){if(ss=='-')f=-1;ss=getchar();}
    while(ss>='0'&&ss<='9'){x=x*10+ss-'0';ss=getchar();}
    return f*x;
}

const int maxn=50;
int n;
struct node{dd x,y,z;}rem[maxn];
dd dx[2]={1.0,-1.0};

dd dist(dd x,dd y,dd z,int k)
{
    dd tx=x-rem[k].x,ty=y-rem[k].y,tz=z-rem[k].z;
    tx*=tx; ty*=ty; tz*=tz;
    return sqrt(tx+ty+tz);
}

int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        if(n==0) break;
        for(int i=1;i<=n;++i)
        scanf("%lf%lf%lf",&rem[i].x,&rem[i].y,&rem[i].z);

        dd t=T,r=100.0,ans=1e9;
        dd x=rem[1].x,y=rem[1].y,z=rem[1].z;
        while(t>eps)
        {
            int mx=1;
            for(int i=1;i<=n;++i)
            if(dist(x,y,z,i)>dist(x,y,z,mx)) mx=i;

            dd dis=dist(x,y,z,mx);
            ans=min(ans,dis);
            x+=(rem[mx].x-x)/dis*t;
            y+=(rem[mx].y-y)/dis*t;
            z+=(rem[mx].z-z)/dis*t;
            t*=delta;
        }
        printf("%.5lf\n",ans);
    }
    return 0;
}