poj 2069

题意：给你n个点，求这n个点的最小外接球的最小半径

淬火法

#include <iostream>
#include <cstdio>
#include <iomanip>
#include <string>
#include <cstdlib>
#include <cstring>
#include <queue>
#include <set>
#include <vector>
#include <map>
#include <algorithm>
#include <cmath>
#include <stack>

#define INF 0x3f3f3f3f
#define LINF 0x3f3f3f3f3f3f3f3f
#define ll long long
#define ull unsigned long long
#define uint unsigned int

using namespace std;

const double eps = 1e-7;
struct point3D {
	double x, y, z;
} a[35];
int n;
double dis(point3D a, point3D 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 step = 100, ans = 1e30, mt;
	point3D z;
	z.x = z.y = z.z = 0;
	int s = 0;
	while (step > eps) {
		for (int i = 0; i < n; i++)
			if (dis(z, a[s]) < dis(z, a[i])) s = i;
		mt = dis(z, a[s]);
		ans = min(ans, mt);
		z.x += (a[s].x - z.x) / mt * step;
		z.y += (a[s].y - z.y) / mt * step;
		z.z += (a[s].z - z.z) / mt * step;
		step *= 0.98;
	}
	return ans;
}
int main(){ // freopen("t.txt","r",stdin);
	double ans;
	while (~scanf("%d", &n), n){
		for (int i = 0; i < n; i++)
			scanf("%lf%lf%lf", &a[i].x, &a[i].y, &a[i].z);
		ans = solve();
		printf("%.5f\n", ans);
	}
	return 0;
}