poj 2420 A Star not a Tree?(시뮬레이션 퇴화 구비 마점)

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
제목:http://poj.org/problem?id=2420
제목: n점을 주고 점 P를 구하세요. 이 점에서 다른 점까지의 거리의 합이 가장 짧습니다.말 좀 써달라고.
분석: 페마점은 아날로그 퇴화로 구할 수 있다. 이 문제는 데이터가 약해서 아무렇게나 지낼 수 있다.
PS: 저는 항상 페마점이 가장 작은 원에 포함된 원심인 줄 알았어요. 그리고 곤란했어요. 아무리 조정해도 지나갈 수가 없어요. 다른 사람의 코드를 봐도 몰라요. 써보고 갑자기 깨달았어요. T_T
코드:

#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
",SimulatedAnnealing(g,n));
    }
    return 0;
}