题目链接：
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 5219 | Accepted: 2491 |
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题目：求一个多边形的费马点到所有点的距离和,费马点是多边形内到顶点距离和最短的点;思路：更poj1379一个人思路,用模拟退火算法找费马点求距离;AC代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <ctime>
#include <algorithm>
using namespace std;
const int num=100;
const int cnt=20;
const double PI=acos(-1.0);
const double inf=1e12;
struct node
{
double px,py,dis;
};
node ans[100];
int n;
double x[105],y[105];
double get_dis(double a,double b,double c,double d)
{
return sqrt((c-a)*(c-a)+(d-b)*(d-b));
}
void Iint()
{
for(int i=0;i<num;i++)
{
ans[i].px=1.0*(rand()%32767)/32767*10000;
ans[i].py=1.0*(rand()%32767)/32767*10000;
ans[i].dis=0;
for(int j=0;j<n;j++)
{
ans[i].dis+=get_dis(ans[i].px,ans[i].py,x[j],y[j]);
}
}
}
int main()
{
srand((unsigned)time(NULL));
scanf("%d",&n);
for(int i=0;i<n;i++)
{
scanf("%lf%lf",&x[i],&y[i]);
}
Iint();
double temp=get_dis(0,0,10000,10000);
while(temp>0.2)
{
for(int i=0;i<num;i++)
{
double fx,fy;
for(int j=0;j<cnt;j++)
{
double cur=2.0*PI*(rand()%32767)/32767;
fx=ans[i].px+cos(cur)*temp;
fy=ans[i].py+sin(cur)*temp;
if(fx<0||fx>10000||fy<0||fy>10000)continue;
double dist=0;
for(int k=0;k<n;k++)
{
dist+=get_dis(fx,fy,x[k],y[k]);
}
if(dist<ans[i].dis)
{
ans[i].px=fx;
ans[i].py=fy;
ans[i].dis=dist;
}
}
}
temp*=0.8;
}
double ans_dis=inf;
int ans_x,ans_y;
for(int i=0;i<num;i++)
{
if(ans[i].dis<ans_dis)
{
ans_dis=ans[i].dis;
}
}
printf("%.0lf\n",ans_dis);
return 0;
}