Brief description :
。广义费马点。。）
Analysis :
。。略。。（。。这个题可以向很多方向推广。。一个比较难的方向是允许放置多个 hub 。。。
const int I = 1, L = 100, W = 65536;
const DB V = 0.9;
// I: 初始点数
// L: 迭代次数
// V: 降火速率。。
const int N = 109, C = 10000;
Po P[N]; int X, Y; Po res; DB ans;
int n;
bool check(Po p){
return p.x >= 0 && p.y >= 0 && p.x <= C && p.y <= C;
}
DB f(Po p){
DB res = 0; REP(i, n) res += dist(P[i], p);
return res;
}
void SA(){
ans = OO; DO_C(I){
Po cur = Po(C / 2, C / 2);
DB T = C / 2, best = f(cur);
while (T > EPS){
bool improved = false;
Po pre = cur; DO_C(L){
DB theta = (DB) (rand() % W) / W * (2*PI);
Po suc = pre + Po(cos(theta), sin(theta)) * T;
DB temp = f(suc);
if (temp < best){
improved = true;
best = temp, cur = suc;
}
}
if (!improved) T *= V;
}
if (best < ans){
ans = best, res = cur;
}
}
}
int main(){
#ifdef LOCAL
freopen("in.txt", "r", stdin);
#endif
while (scanf("%d", &n) != EOF && n){
REP(i, n) P[i].input();
SA(); OT(ans);
}
}
