poj 2420 A Star not a Tree?

题意：给出平面上N（<=100）个点，你需要找到一个这样的点，使得这个点到N个点的距离之和尽可能小。输出这个最小的距离和（四舍五入到最近的整数）

思路：三分求

三分算法解决凸形或者凹形函数的极值；

二分解决具有单调性的函数的极值；

如下图

[code]mid = （l+r） / 2;
midmid = (r+mid)  / 2;
求极大值
if(cal(mid)>cal(midmid))  r= midmid;
else l = mid;

求极小值
if(cal(mid)<cal(midmid))  r= midmid;
else l = mid;

[code]/* **********************************************
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
;

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));
}