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. QTextEdit中选中文本修改字体与颜色,全部文本修改字体与颜色(设置调色板的前景色、背景色、文字颜色以及基色)

    ----我的生活,我的点点滴滴!! 当然以下内容都可以通过设置样式来达到目的,但是下面不使用这样的方法 先来看张图,理解此图基本就能实现上面所要达到的目的了 Widget::Widget(QWidge ...

  2. AIX安装SSH

    AIX 7.1   进入 https://www14.software.ibm.com/webapp/iwm/web/preLogin.do?source=aixbp 先下载需要的openssl,op ...

  3. 关于linux系统下的uname -a命令

  4. iOS 创建推送证书

    1.首先你想创建推送证书和以前你做真机测试证书一样,需要实现准备一个99$的付费账号.然后登陆苹果开发者网站.http://developer.apple.com/ 2.登陆以后你能看到这个界面然后选 ...

  5. 最少步数(dfs + bfs +bfs优化)

    最少步数 时间限制:3000 ms  |  内存限制:65535 KB 难度:4   描述 这有一个迷宫,有0~8行和0~8列: 1,1,1,1,1,1,1,1,1 1,0,0,1,0,0,1,0,1 ...

  6. spring 3配置文件中如何注入map list set等类型

    首先写个 javabean类吧,如下 package com.bean; import java.util.List; import java.util.Map; import java.util.P ...

  7. IKAnalyzer使用停用词词典进行分词

    @Test // 測试分词的效果,以及停用词典是否起作用 public void test() throws IOException { String text = "老爹我们都爱您.&qu ...

  8. 【main()的参数探究】

    恩...今天研究信安的课件的时候看到一段对于main(int argc,char *argv[])的编程 所以探究探究main()函数的参数 探究程序如下: #include <cstdio&g ...

  9. SQL Server Reporting Services (SQLEXPRESS) 服务占用80端口

    win7, 好多时候,看到system进程占用了80端口,这个是系统进程,不能直接结束.我们不知道这个进程的哪个服务占用了80端口,这里记录其中一个服务"SQL Server Reporti ...

  10. 每日必读(2) --Base64

    一. base64是什么? 按照RFC2045的定义,Base64被定义为:Base64内容传送编码被设计用来把任意序列的8位字节描述为一种不易被人直接识别的形式.(The Base64 Conten ...