[gym101981D][2018ICPC南京D题]Country Meow

题目链接

题目大意是求三维空间可以包含$n$个点的最小圆半径。

如果有做过洛谷P1337就会发现这到题很模拟退火，所以就瞎搞一发。

$PS:$注意本题时限$3$秒。

 #include<bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 const int maxn = ;
 struct node {
     double x, y, z;
 }a[maxn];
 int n;
 double ansx, ansy, ansz, ans;
 double dis(double x, double y, double z, node w) {
     return sqrt((x - w.x) * (x - w.x) + (y - w.y) * (y - w.y) + (z - w.z) * (z - w.z));
 }
 double solve(double x, double y, double z) {
     double w = ;
     for (int i = ; i <= n; i++)
         w = max(w, dis(x, y, z, a[i]));
     return w;
 }
 const double delta = 0.998;
 void SA() {
     double X = ansx, Y = ansy, Z = ansz;
     double t = ;
     while (t > 1e-) {
         double x = X + (rand() *  - RAND_MAX) * t;
         double y = Y + (rand() *  - RAND_MAX) * t;
         double z = Z + (rand() *  - RAND_MAX) * t;
         double now = solve(x, y, z);
         double D = now - ans;
         if (D < ) {
             X = x, Y = y, Z = z;
             ansx = X, ansy = Y;
             ansz = Z;
             ans = now;
         }
         else if (exp(-D / t) * RAND_MAX > rand())X = x, Y = y, Z = z;
         t *= delta;
     }
 }
 int main() {
     srand(unsigned(time(NULL)));
     scanf("%d", &n);
     for (int i = ; i <= n; i++) {
         scanf("%lf%lf%lf", &a[i].x, &a[i].y, &a[i].z);
         ansx += a[i].x, ansy += a[i].y, ansz += a[i].z;
     }
     ansx /= n, ansy /= n, ansz /= n;
     ans = solve(ansx, ansy, ansz);
     while ((double)clock() / CLOCKS_PER_SEC < 2.7)
         SA();
     printf("%.10f", ans);
 }