A Star not a Tree?
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 4058   Accepted: 2005

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
 
模拟退火算法、网上各种写法、晕死、都不知道哪个是对的了、
数据弱、乱写的、AC了、
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <ctime>
#include <cstdlib>
using namespace std;
#define INF 1e20
#define PI acos(-1.0)
#define N 110 struct Point
{
double x,y;
Point(){}
Point(double x,double y):x(x),y(y){}
};
int n;
Point p[N]; double cal(Point t)
{
double sum=;
for(int i=;i<=n;i++) sum+=sqrt((p[i].x-t.x)*(p[i].x-t.x)+(p[i].y-t.y)*(p[i].y-t.y));
return sum;
}
void solve()
{
int TN=,DN=;
Point u,v,ansp;
double ud,vd,ansd=INF;
double step=,eps=1e-,r=0.85; while(TN--)
{
u=Point(rand()%,rand()%);
ud=cal(u);
while(step>eps)
{
bool flag=;
while(flag)
{
flag=;
for(int i=;i<DN;i++)
{
double d=*PI*(double)rand()/RAND_MAX;
v.x=u.x+sin(d)*step;
v.y=u.y+cos(d)*step;
vd=cal(v);
if(vd<ud) ud=vd,u=v,flag=;
}
}
step*=r;
}
if(ud<ansd) ansd=ud,ansp=u;
}
printf("%.0f\n",ansd);
}
int main()
{
//srand((int)time(NULL)); //POJ上不能用
while(scanf("%d",&n)!=EOF)
{
for(int i=;i<=n;i++) scanf("%lf%lf",&p[i].x,&p[i].y);
solve();
}
return ;
}

[POJ 2420] A Star not a Tree?的更多相关文章

  1. 三分 POJ 2420 A Star not a Tree?

    题目传送门 /* 题意:求费马点 三分:对x轴和y轴求极值,使到每个点的距离和最小 */ #include <cstdio> #include <algorithm> #inc ...

  2. POJ 2420 A Star not a Tree? 爬山算法

    B - A Star not a Tree? Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hust.edu.cn/vjudge/co ...

  3. POJ 2420 A Star not a Tree? (计算几何-费马点)

    A Star not a Tree? Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 3435   Accepted: 172 ...

  4. poj 2420 A Star not a Tree?——模拟退火

    题目:http://poj.org/problem?id=2420 精度设成1e-17,做三遍.ans设成double,最后再取整. #include<iostream> #include ...

  5. poj 2420 A Star not a Tree? —— 模拟退火

    题目:http://poj.org/problem?id=2420 给出 n 个点的坐标,求费马点: 上模拟退火. 代码如下: #include<iostream> #include< ...

  6. POJ 2420 A Star not a Tree?(模拟退火)

    题目链接 居然1Y了,以前写的模拟退火很靠谱啊. #include <cstdio> #include <cstring> #include <string> #i ...

  7. POJ 2420 A Star not a Tree?【爬山法】

    题目大意:在二维平面上找出一个点,使它到所有给定点的距离和最小,距离定义为欧氏距离,求这个最小的距离和是多少(结果需要四舍五入)? 思路:如果不能加点,问所有点距离和的最小值那就是经典的MST,如果只 ...

  8. 【POJ】2420 A Star not a Tree?(模拟退火)

    题目 传送门:QWQ 分析 军训完状态不好QwQ,做不动难题,于是就学了下模拟退火. 之前一直以为是个非常nb的东西,主要原因可能是差不多省选前我试着学一下但是根本看不懂? 骗分利器,但据说由于调参困 ...

  9. 【POJ】2420 A Star not a Tree?

    http://poj.org/problem?id=2420 题意:给n个点,求一个点使得到这个n个点的距离和最短,输出这个最短距离(n<=100) #include <cstdio> ...

随机推荐

  1. 这个SpringMVC的一直刷屏的问题你见过吗?无解

    严重: Servlet.service() for servlet DispatcherServlet threw exceptionjava.lang.StackOverflowError at o ...

  2. SQL动态更新表字段 传入字段可能为空

    小技巧: 项目组有修改产品的基本信息字段 但有时候传入的字段可能为空 也可能不为空  动态修改表中字段. USE [BetaProductMarket_DB] GO )) BEGIN DROP PRO ...

  3. 浅析nginx的负载均衡

    Nginx 的 HttpUpstreamModule 提供对后端(backend)服务器的简单负载均衡.一个最简单的 upstream 写法如下: upstream backend { server ...

  4. PHP微信开发代码

    1,SAE上申请服务器 2,绑定测试账号 3,Token验证 <?php /* http://www.cnblogs.com/xrhou12326/ CopyRight 2014 All Rig ...

  5. Hibernate从入门到精通(七)多对一单向关联映射

    上次的博文Hibernate从入门到精通(六)一对一双向关联映射中我们介绍了一下一对一双向关联映射,本次博文我们讲解一下多对一关联映射 多对一单向关联映射 多对一关联映射与一对一关联映射类似,只是在多 ...

  6. jsf2.0视频

    jsf2.0 入门视频 教程   需要的看下.初次录视频.还有很多需要完善. JSF交流QQ群84376982 JSF入门视频下载地址  http://pan.baidu.com/s/1jG3y4T4 ...

  7. 机器学习基石的泛化理论及VC维部分整理

    第四讲 机器学习的可行性 一.Hoeffding's Inequality \(P[\left | \nu -\mu  \right |>\epsilon ] \leq 2exp(-2\epsi ...

  8. js页面刷新一次

    // var str = document.location.hash, // index = str.indexOf("#"); // if(index == 0){ // wi ...

  9. jQuery树结构插件推荐zTree

    JQuery zTree 下载地址http://plugins.jquery.com/zTree.v3/

  10. firefly的环境搭建(2013年9月25日最新,win下最详图文)

    源地址:http://www.9miao.com/question-15-53785.html 一.安装PythonFirefly是采用Python编写的高性能.分布式游戏服务器框架,所以使用Fire ...