題意：給出平面上N（<=100）個點，你需要找到一個這樣的點，使得這個點到N個點的距離之和儘可能小。輸出這個最小的距離和（四捨五入到最近的整數）

思路：三分求

三分演算法解決凸形或者凹形函式的極值；
二分解決具有單調性的函式的極值；
如下圖

mid = （l+r） / 2;
midmid = (r+mid)  / 2;
求極大值
if(cal(mid)>cal(midmid))  r= midmid;
else l = mid;

求極小值
if(cal(mid)<cal(midmid))  r= midmid;
else l = mid;

/* **********************************************
Auther: xueaohui
Created Time: 2015-7-25 8:56:03
File Name   : poj2420.cpp
*********************************************** */
 

#include <iostream>
#include <fstream>
#include <cstring>
#include <climits>
#include <deque>
#include <cmath>
#include <queue>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <utility>
#include <sstream>
 

#include <complex>
#include <string>
#include <vector>
#include <cstdlib>
#include <cstdio>
#include <ctime>
#include <bitset>
#include <functional>
#include <algorithm>
using namespace std;
#define ll long long
#define N 111

int n;

struct node{
    double
 
 x,y;
    double dis(node a){
        return sqrt((x-a.x)*(x-a.x)+(y-a.y)*(y-a.y));
    }
}p[N];

double sum(double xx,double yy){
    node a;
    a.x = xx;
    a.y = yy;
    double d=0;
    for(int i=1;i<=n;i++){
        d+=a.dis(p[i]);
    }
    return d;
}

double cal(double xx){
    double l = 0;
    double r = 10000;
    for(int i=1;i<=200;i++){
        double mid = (l+r)/2;
        double rmid = (mid+r)/2;
        if(sum(xx,mid)<sum(xx,rmid)){
            r = rmid;
        }
        else{
            l = mid;
        }
    }
    return sum(xx,l);
}

int main(){
    scanf("%d",&n);
    for(int i=1;i<=n;i++){
        scanf("%lf%lf",&p[i].x,&p[i].y);
    }
    double l = 0;
    double r = 10000;
    for(int i=1;i<=200;i++){
        double mid = (l+r)/2;
        double rmid = (mid+r)/2;
        if(cal(mid)<cal(rmid)){
           r = rmid;
        }
        else{
            l = mid;
        }
    }
    printf("%.0f\n",cal(l));
}