Network

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 60000/30000K (Java/Other)
Total Submission(s) : 17   Accepted Submission(s) : 8
Special Judge
Problem Description
Andrew is working as system administrator and is planning to establish a new network in his company. There will be N hubs in the company, they can be connected to each other using cables. Since each worker of the company must have access to the whole network, each hub must be accessible by cables from any other hub (with possibly some intermediate hubs).
Since cables of different types are available and shorter ones are cheaper, it is necessary to make such a plan of hub connection, that the maximum length of a single cable is minimal. There is another problem not each hub can be connected to any other one because of compatibility problems and building geometry limitations. Of course, Andrew will provide you all necessary information about possible hub connections.
You are to help Andrew to find the way to connect hubs so that all above conditions are satisfied.
 
Input
The first line of the input contains two integer numbers: N - the number of hubs in the network (2 <= N <= 1000) and M - the number of possible hub connections (1 <= M <= 15000). All hubs are numbered from 1 to N. The following M lines contain information about possible connections - the numbers of two hubs, which can be connected and the cable length required to connect them. Length is a positive integer number that does not exceed 106. There will be no more than one way to connect two hubs. A hub cannot be connected to itself. There will always be at least one way to connect all hubs.
 
Output
Output first the maximum length of a single cable in your hub connection plan (the value you should minimize). Then output your plan: first output P - the number of cables used, then output P pairs of integer numbers - numbers of hubs connected by the corresponding cable. Separate numbers by spaces and/or line breaks.
 
Sample Input
4 6 1 2 1 1 3 1 1 4 2 2 3 1 3 4 1 2 4 1
 
Sample Output
1 4 1 2 1 3 2 3 3 4
 题解:
一:找最小生成树的最小权值;
二:生成这个数的数据组数;
三:输出这些数据;
这题样例是错的,无语;
代码:
 #include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define MAX(x,y) (x>y?x:y)
const int MAXN=;
struct Node{
int s,e,c;
};
int cmp(Node a,Node b){
return a.c<b.c;
}
Node dt[MAXN];
int pre[];
int pt[MAXN],k,flot,as;
int find(int x){
return pre[x]= x==pre[x]?x:find(pre[x]);
}
void add(int a,int b){
pt[k++]=a;pt[k++]=b;
}
void initial(){
memset(pre,,sizeof(pre));
k=;flot=;
as=-;
}
void merge(Node a){
int f1,f2;
if(!pre[a.s])pre[a.s]=a.s;
if(!pre[a.e])pre[a.e]=a.e;
f1=find(a.s);f2=find(a.e);
if(f1!=f2){
add(a.s,a.e);
flot++;
as=MAX(as,a.c);
pre[f1]=f2;
}
}
int main(){int N,M;
while(~scanf("%d%d",&N,&M)){
initial();
for(int i=;i<M;i++){
scanf("%d%d%d",&dt[i].s,&dt[i].e,&dt[i].c);
}
sort(dt,dt+M,cmp);
for(int i=;i<M;i++){
merge(dt[i]);
}
printf("%d\n%d\n",as,k/);
for(int i=;i<k;i+=){
printf("%d %d\n",pt[i],pt[i+]);
}
}
return ;
}

Network()的更多相关文章

  1. [转]VMware虚拟机上网络连接(network type)的三种模式--bridged、host-only、NAT

    转自:http://www.cnblogs.com/xiaochaohuashengmi/archive/2011/03/15/1985084.html VMWare提供了三种工作模式,它们是brid ...

  2. VMware虚拟机上网络连接(network type)的三种模式--bridged、host-only、NAT

    VMware虚拟机上网络连接(network type)的三种模式--bridged.host-only.NAT VMWare提供了三种工作模式,它们是bridged(桥接模式).NAT(网络地址转换 ...

  3. POJ 3659 Cell Phone Network / HUST 1036 Cell Phone Network(最小支配集,树型动态规划,贪心)-动态规划做法

    POJ 3659 Cell Phone Network / HUST 1036 Cell Phone Network(最小支配集,树型动态规划,贪心) Description Farmer John ...

  4. 【POJ 3694】 Network(割边&lt;桥&gt;+LCA)

    [POJ 3694] Network(割边+LCA) Network Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 7971 ...

  5. 更换oracle 集群网卡(Changing a Network Interface)

     更换oracle 集群网卡(Changing a Network Interface) 假设换网卡前后 网卡名.ip,网关,子网掩码都不变的话,集群层面不许要做额外的操作. 一下操作为更换网卡后 ...

  6. Efficient and Accurate Arbitrary-Shaped Text Detection with Pixel Aggregation Network(利用像素聚合网络进行高效准确的任意形状文本检测)

    PSENet V2昨日刚出,今天翻译学习一下. 场景文本检测是场景文本阅读系统的重要一步,随着卷积神经网络的快速发展,场景文字检测也取得了巨大的进步.尽管如此,仍存在两个主要挑战,它们阻碍文字检测部署 ...

  7. ZOJ - 1586 QS Network (Prim)

    ZOJ - 1586 QS Network (Prim) #include<iostream> #include<cstring> using namespace std; + ...

  8. aborb()程序结束形式

    abort()与exit()的区别?         分类:             MFC              2011-01-04 14:13     2233人阅读     评论(0)   ...

  9. Select()使用否?

    David Treadwell ,Windows Socket 的一位开发者,曾经在他的一篇名为"Developing Transport-Independent Applications ...

随机推荐

  1. Android下EditText中的字体不统一问题

    Android下EditText中的字体不统一问题 好久没写,今天心情好略记下解决的某bug 在一个登录界面有帐号和密码两个EditText,但是却发现两个EditText的hint的英文字体不同,看 ...

  2. USB Mass Storage大容量存储的基本知识

    http://www.crifan.com/files/doc/docbook/usb_disk_driver/release/htmls/ch02_msc_basic.html 目录 2.1. US ...

  3. 首次登录与在线求助man page

    为了避免瞬间断电造成的Linux系统损害,建议作为服务器的Linux主机应该加上不断电系统来持续提供稳定的电力. 在终端环境中,可依据提示符为$或#判断为一般几号或root账号. 要取得终端支持的语言 ...

  4. android 滚动条

    ScrollView简单应用 activity中经常只是一个LinearLayout,但这样的话,如果activity内容超过一屏,无法滚动查看下面的内容. 这时只需在外面嵌套一个ScrollView ...

  5. C#接口的使用【转】

    我们定义一个接口public interface IBark{   void Bark();}再定义一个类,继承于IBark,并且必需实现其中的Bark()方法public class Dog:IBa ...

  6. 张冬:OpenPOWER CAPI为什么这么快?(二)

     张冬:OpenPOWER CAPI为什么这么快?(二) PMC公司数据中心存储架构师张冬 有了CAPI的FPGA是怎么做的? 首先认识一下这个体系里的三个角色: AFU(Acceleration ...

  7. POJ 3468 A Simple Problem with Integers(线段树区间求和)

    Description You have N integers, A1, A2, ... , AN. You need to deal with two kinds of operations. On ...

  8. JS实现文本复制与剪切

    我们在网页上放置一个复制按钮,主要用来方便用户复制链接之类的复杂文本,以往的做法是,通过JS依靠Flash,甚至借助jQuery庞大的js库来实现文本复制到剪贴板的.今天我要给大家介绍的是一款极现代的 ...

  9. 移动端的几款jq插件

    移动手机用户的数量每日都在增长,人们现在都习惯于使用手机来浏览网页,看小说,读新闻.如何确保你的网站对移动用户友好,是目前你需要解决的最重要的问 题之一.这里给大家介绍10款在移动手机上使用的jQue ...

  10. UVA1600 Patrol Robot

    题意: 求机器人走最短路线,而且可以穿越障碍.N代表有N行,M代表最多能一次跨过多少个障碍. 分析: bfs()搜索,把访问状态数组改成了3维的,加了个维是当前能跨过的障碍数. 代码: #includ ...