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.

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.

#include<iostream>
#include<cmath>
#include<ctime>
#include<cstdio>
using namespace std;

const int MAXN=50;
const double eps=1e-8;
int n;

struct Point
{
	double x,y,z;
}myPoint[MAXN],op;

double GetDis(const Point &a,const 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 ret,delta=100.0;
	double maxDis,tempDis;
	int i,id;
	while(delta>eps)
	{
		id=0;
		maxDis=GetDis(op,myPoint[id]);
		for(i=1;i<n;i++)
		{
			tempDis=GetDis(op,myPoint[i]);
			if(tempDis>maxDis)
			{
				maxDis=tempDis;
				id=i;
			}
		}
		ret=maxDis;
		op.x+=(myPoint[id].x-op.x)/maxDis*delta;
		op.y+=(myPoint[id].y-op.y)/maxDis*delta;
		op.z+=(myPoint[id].z-op.z)/maxDis*delta;
		delta*=0.98;
	}
	return ret;
}

int main()
{
	int i;
	while(scanf("%d",&n)!=EOF)
	{
		if(0==n)break;
		for(i=0;i<n;i++)
			scanf("%lf%lf%lf",&myPoint[i].x,&myPoint[i].y,&myPoint[i].z);
		op.x=op.y=op.z=0.0;

		printf("%.5lf\n",Solve());
	}
	return 0;
}

Buried memory

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2213    Accepted Submission(s): 1212

3
1.00 1.00
2.00 2.00
3.00 3.00
0

2.00 2.00 1.41

#include<iostream>
#include<cmath>
#include<ctime>
#include<cstdio>
using namespace std;

const int MAXN=500+10;
const double eps=1e-8;
int n;

struct Point
{
	double x,y;
}myPoint[MAXN],op;

double GetDis(const Point &a,const Point &b)
{
	return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}

void Solve()
{
	double ret,delta=100.0;
	double maxDis,tempDis;
	int i,id;
	while(delta>eps)
	{
		id=0;
		maxDis=GetDis(op,myPoint[id]);
		for(i=1;i<n;i++)
		{
			tempDis=GetDis(op,myPoint[i]);
			if(tempDis>maxDis)
			{
				maxDis=tempDis;
				id=i;
			}
		}
		ret=maxDis;
		op.x+=(myPoint[id].x-op.x)/maxDis*delta;
		op.y+=(myPoint[id].y-op.y)/maxDis*delta;
		delta*=0.98;
	}
	printf("%.2lf %.2lf %.2lf\n",op.x,op.y,ret);
}

int main()
{
	int i;
	while(scanf("%d",&n)!=EOF)
	{
		if(0==n)break;
		op.x=op.y=0;
		for(i=0;i<n;i++)
		{
			scanf("%lf%lf",&myPoint[i].x,&myPoint[i].y);
			op.x+=myPoint[i].x;
			op.y+=myPoint[i].y;
		}
		op.x/=n;
		op.x/=n;
		Solve();
	}
	return 0;
}