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. 转 在无法通过yum下载非标准包时,怎么办

    在CentOS下,我们可以通过yum来下载或更新rpm包,但是标准的源(repository)里只提供一部分的rpm包,虽然大部分情况下,这些包是够用的.但是有时候还是需要下载其他的一些非标准的包,如 ...

  2. c# 代理IP获取通用方法

    调用: ConcurrentQueue<string> proxyIpQueue = new ConcurrentQueue<string>(); Grab_ProxyIp(p ...

  3. c语言随机数

    写得我自己都看不好了:大家都比较喜欢吃快餐,只需要尽快告诉读者怎么用起来就行了.不想听啰啰嗦嗦说一堆,然后例程还特别麻烦 so:  1.基本 int seed = time(0);//#include ...

  4. 图像处理-07-图像的轮廓提取-Robert算子

    图像的轮廓提取-Robert算子 图像的边缘:周围像素灰度有阶跃变化或“屋顶”变化的那些像素的集合,边缘广泛存在于物体与背景之间.物体与物体之间,基元与基元之间,是图像分割的重要依据. 物体的边缘是由 ...

  5. 【js】IE、FF、Chrome浏览器中的JS差异介绍

    如何判断浏览器类型 转:http://www.cnblogs.com/carekee/articles/1854674.html 1.通过浏览器特有的对象 如ie 的ActiveXObject  ff ...

  6. python学习笔记8(表达式和语句)

    一.print 和 import 信息 >>> print 'age:',22 # 用逗号隔开打印多个表达式 age: 22 import somemodule # 从模块导入函数 ...

  7. (转)Linux 文件系统:procfs, sysfs, debugfs 用法简介

    网址:http://www.tinylab.org/show-the-usage-of-procfs-sysfs-debugfs/ 1 前言 内核中有三个常用的伪文件系统:procfs,debugfs ...

  8. linux编程之指针

    这个是数组指针.指针数组.二维数组之间相互转换的代码 #include<stdio.h> void main() { ][]={,,,,,,,}; int *b=NULL; int **c ...

  9. spring的配置模式与注解模式基础

    “依赖注入”是spring的核心特征,在Web服务器(如Tomcat)加载时,它会根据Spring的配置文件中配置的bean或者是通过注解模式而扫描并装载的bean实例自动注入到Application ...

  10. 在TNSNAMES.ORA文件中配置本机装的oracle

    首先,感谢这两位网友:http://zhidao.baidu.com/link?url=eGYeoEa-EhQdVitSGqjE36uNfVmEsryXH1WUjPue6YvArDSx-Y1N9_rd ...