zoukankan      html  css  js  c++  java
  • POJ1861 Network

    Time Limit: 1000MS   Memory Limit: 30000KB   64bit IO Format: %lld & %llu

    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 10 6. 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
    

    样例输出明显错了。

    1

    3

    1 2

    1 3

    3 4

    这样可行。有Special Judge

     

    Source

    Northeastern Europe 2001, Northern Subregion
     
     
    裸最小生成树。要求输出生成树中最大边数,连接的点数(肯定是n-1),使用的每条边。
     
     1 /*by SilverN*/
     2 #include<algorithm>
     3 #include<iostream>
     4 #include<cstring>
     5 #include<cstdio>
     6 #include<cmath>
     7 #include<queue>
     8 using namespace std;
     9 int read(){
    10     int x=0,f=1;char ch=getchar();
    11     while(ch<'0' || ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    12     while(ch>='0' && ch<='9'){x=x*10+ch-'0';ch=getchar();}
    13     return x*f;
    14 }
    15 const int mxn=16000;
    16 int n,m;
    17 struct node{
    18     int x,y;
    19     int dis;
    20 }e[mxn];
    21 int cnt=0;
    22 int cmp(node a,node b){
    23     return a.dis<b.dis;
    24 }
    25 int fa[mxn];
    26 void init(){
    27     for(int i=1;i<=n;i++)fa[i]=i;
    28 }
    29 int find(int x){
    30     if(fa[x]==x)return x;
    31     return fa[x]=find(fa[x]);
    32 }
    33 queue<int>aw;
    34 int ans=0;
    35 void Kruskal(){
    36     int i,j;
    37     ans=0;
    38     int now=0;
    39     for(i=1;i<=m;i++){
    40         int u=find(e[i].x),v=find(e[i].y);
    41         if(u==v)continue;
    42         aw.push(i);
    43         fa[u]=v;
    44         now++;
    45         ans=max(ans,e[i].dis);
    46         if(now==n-1)break;
    47     }
    48 }
    49 int main(){
    50     while(scanf("%d",&n)!=EOF){
    51         init();
    52         int u,v,dis;
    53         m=read();
    54         int i,j;
    55         for(i=1;i<=m;i++){
    56             e[i].x=read();e[i].y=read();e[i].dis=read();
    57         }
    58         sort(e+1,e+m+1,cmp);
    59         Kruskal();
    60         printf("%d
    ",ans);
    61         printf("%d
    ",n-1);
    62         while(!aw.empty()){
    63             int tmp=aw.front();
    64             aw.pop();
    65             printf("%d %d
    ",e[tmp].x,e[tmp].y);
    66         }
    67     }
    68     return 0;
    69 }
  • 相关阅读:
    解决:信息中插入avi格式的视频时,提示“unsupported video format”
    java字节数组和16进制之间的转换
    16进制转换字节数组工具类
    如何在ubuntu 12.04 中安装经典的 GNOME桌面
    Ubuntu安装软件提示”需要安装不能信任的软件包”解决办法
    Ubuntu系统下运行Eclipse出现找不到jre的问题的解决方法
    ubuntu添加共享出错
    从scrapy使用经历说开来
    有趣的问题--12 coins problem
    一个奇怪的MySQL错误返回
  • 原文地址:https://www.cnblogs.com/SilverNebula/p/5892664.html
Copyright © 2011-2022 走看看