The 2018 ICPC Asia Nanjing Regional D - Country Meow 模拟退火

D - Country Meow

solution

最小球覆盖

code

/*Siberian Squirrel*/
/*Cute KiloFish*/
#include<bits/stdc++.h>

#define IO ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)

using namespace std;
typedef long long ll;

const double PI = acos(-1);
const double eps = 1e-4;
const int INF = 0x3f3f3f3f;
/*const int MOD = 998244353, r = 119, k = 23, g = 3;
const int MOD = 1004535809, r = 479, k = 21, g = 3;*/
const int MOD = 1e9 + 7;
const int M = 1e7 + 10;
const int N = 2e5 + 10;

//inline int rnd(){static int seed=2333;return seed=(((seed*666666ll+20050818)%998244353)^1000000007)%1004535809;}

int n, pos;
struct node {
    
    
    double x, y, z;
}p[N], _p;
double maxx = 0;

double dis(node a, node 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));
}
void solve(double res = 1.0 * INF) {
    
    
    double t = 10000.0;
    while (t > eps) {
    
    
        maxx = 0;
        for (int i = 1; i <= n; ++ i) {
    
    
            if (dis(_p, p[i]) > maxx) {
    
    
                maxx = dis(_p, p[i]);
                pos = i;
            }
        }
        res = min(res, maxx);
        _p.x += (p[pos].x - _p.x) * (t / 10000.0);
        _p.y += (p[pos].y - _p.y) * (t / 10000.0);
        _p.z += (p[pos].z - _p.z) * (t / 10000.0);
        t *= 0.98;
    }
    cout << fixed << setprecision(19) << res << endl;;
}

signed main() {
    
    
    IO;
#ifdef ACM_LOCAL
    freopen("input", "r", stdin);
    freopen("output", "w", stdout);
#endif
    int o = 1, cases = 0;
//    cin >> o;
    while(o --) {
    
    
        cin >> n;
        for(int i=1; i<=n; ++i) {
    
    
            cin >> p[i].x >> p[i].y >> p[i].z;
        }
        solve();
    }
    return 0;
}