poj 2420 A Star not a Tree?(模拟退火求费马点)

A Star not a Tree? Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 2519   Accepted: 1332 Description Luke wants to upgrade his home computer network from 10mbs to 100mbs. His existing network uses 10base2 (coaxial) cables that allow you to connect any number of computers together in a linear arrangement. Luke is particulary proud that he solved a nasty NP-complete problem in order to minimize the total cable length.  Unfortunately, Luke cannot use his existing cabling. The 100mbs system uses 100baseT (twisted pair) cables. Each 100baseT cable connects only two devices: either two network cards or a network card and a hub. (A hub is an electronic device that interconnects several cables.) Luke has a choice: He can buy 2N-2 network cards and connect his N computers together by inserting one or more cards into each computer and connecting them all together. Or he can buy N network cards and a hub and connect each of his N computers to the hub. The first approach would require that Luke configure his operating system to forward network traffic. However, with the installation of Winux 2007.2, Luke discovered that network forwarding no longer worked. He couldn't figure out how to re-enable forwarding, and he had never heard of Prim or Kruskal, so he settled on the second approach: N network cards and a hub.  Luke lives in a loft and so is prepared to run the cables and place the hub anywhere. But he won't move his computers. He wants to minimize the total length of cable he must buy. Input The first line of input contains a positive integer N <= 100, the number of computers. N lines follow; each gives the (x,y) coordinates (in mm.) of a computer within the room. All coordinates are integers between 0 and 10,000. Output Output consists of one number, the total length of the cable segments, rounded to the nearest mm. Sample Input 4 0 0 0 10000 10000 10000 10000 0 Sample Output 28284 Source Waterloo Local 2002.01.26

题目：http://poj.org/problem?id=2420

题意：给你n个点，求一个点P，该点到其他点的距离之和最短。也就是求费马点啦

分析：费马点可以用模拟退火来求，这题数据弱，随便都能过的= =

PS：我一直以为费马点就是最小圆包含的圆心，然后囧了，怎么调都过不了，看别人的代码，还是不知道，写过了才突然明白过来 T_T

代码：

#include<cmath>
#include<cstdio>
#include<iostream>
using namespace std;
const int mm=111;
const double eps=1e-8;
int dx[]={0,0,-1,1};
int dy[]={-1,1,0,0};
typedef double diy;
struct point
{
    diy x,y;
    point(){}
    point(diy _x,diy _y):x(_x),y(_y){}
}g[mm];
double Dis(point P,point Q)
{
    return sqrt((P.x-Q.x)*(P.x-Q.x)+(P.y-Q.y)*(P.y-Q.y));
}
double GetAll(point *g,point o,int n)
{
    double ret=0;
    while(n--)ret+=Dis(g[n],o);
    return ret;
}
double SimulatedAnnealing(point *g,int n)
{
    point o=g[0],p;
    double t,tmp,ret=1e99;
    int i,flag;
    for(t=100;t>eps;t*=0.98)
    {
        flag=1;
        while(flag)
        {
            flag=0;
            for(i=0;i<4;++i)
            {
                p.x=o.x+dx[i]*t;
                p.y=o.y+dy[i]*t;
                if(ret>GetAll(g,p,n))
                {
                    ret=GetAll(g,p,n);
                    o=p;
                    flag=1;
                }
            }
        }
    }
    return ret;
}
int main()
{
    int i,n;
    while(~scanf("%d",&n))
    {
        for(i=0;i<n;++i)
            scanf("%lf%lf",&g[i].x,&g[i].y);
        printf("%.0lf\n",SimulatedAnnealing(g,n));
    }
    return 0;
}