poj 2420 模拟退火+费马点

/*
题意：给出N台电脑的位置，找出某个点到N个电脑的距离之和最小

题解：费马点+模拟退火(模板题)
模板题用模拟退火求费马点
*/
#include <iostream>
#include <cmath>

using namespace std;

struct point
{
    double x,y;
}p[105];

int dir[8][2] = {-1,-1,-1,0,-1,1,0,-1,0,1,1,-1,1,0,1,1};

double getdis(point a, point b)
{
    return sqrt((a.x-b.x)*(a.x-b.x) + (a.y-b.y)*(a.y-b.y));
}

double allDis(int n , point f)
{
    double sum = 0;
    for(int i = 0 ; i < n ; i++)
        sum += getdis(p[i],f);
    return sum;
}

point fermat(int n)
{
    double step = 0;
    for (int i = 0 ; i < n ; i++)
        step += fabs(p[i].x) + fabs(p[i].y);

    point f;
    f.x = 0;
    f.y = 0;
    for (int i = 0 ; i < n ; i++)
        f.x += p[i].x , f.y +=p[i].y;
    f.x /= n;
    f.y /= n;
    point t;
    while(step > 1e-10)
    {
        for (int i = 0 ; i < 8 ; i++)
        {
            t.x = f.x + dir[i][0]*step;
            t.y = f.y + dir[i][1]*step;
            if(allDis(n,t) < allDis(n,f))
                f = t;
        }
        step *=0.7;  //步长改动
    }
    return f;
}

int main(void)
{
    int n;
    while (cin >> n)
    {
        for(int i=0; i<n; i++)
            cin >> p[i].x >> p[i].y;
        double ans = allDis(n, fermat(n));
        int t = ans*10;
        if (t%10 < 5)
            cout << t/10 << endl;
        else
            cout << t/10+1 << endl;
    }
    return 0;
}