zoukankan      html  css  js  c++  java
  • POJ 1144 Network

    Network

     

    Description

    A Telephone Line Company (TLC) is establishing a new telephone cable network. They are connecting several places numbered by integers from 1 to N . No two places have the same number. The lines are bidirectional and always connect together two places and in each place the lines end in a telephone exchange. There is one telephone exchange in each place. From each place it is 
    possible to reach through lines every other place, however it need not be a direct connection, it can go through several exchanges. From time to time the power supply fails at a place and then the exchange does not operate. The officials from TLC realized that in such a case it can happen that besides the fact that the place with the failure is unreachable, this can also cause that some other places cannot connect to each other. In such a case we will say the place (where the failure 
    occured) is critical. Now the officials are trying to write a program for finding the number of all such critical places. Help them.

    Input

    The input file consists of several blocks of lines. Each block describes one network. In the first line of each block there is the number of places N < 100. Each of the next at most N lines contains the number of a place followed by the numbers of some places to which there is a direct line from this place. These at most N lines completely describe the network, i.e., each direct connection of two places in the network is contained at least in one row. All numbers in one line are separated 
    by one space. Each block ends with a line containing just 0. The last block has only one line with N = 0;

    Output

    The output contains for each block except the last in the input file one line containing the number of critical places.

    Sample Input

    5
    5 1 2 3 4
    0
    6
    2 1 3
    5 4 6 2
    0
    0

    Sample Output

    1
    2

    Hint

    You need to determine the end of one line.In order to make it's easy to determine,there are no extra blank before the end of each line.
     1 #include<cstdio>
     2 #include<cstring>
     3 #include<algorithm>
     4 #include<vector>
     5 using namespace std;
     6 
     7 vector<int>G[105];
     8 int dfs_clock;
     9 int pre[105],low[105];
    10 bool iscut[105];
    11 int N;
    12 
    13 void init()
    14 {
    15     memset(low,0,sizeof(low));
    16     memset(pre,0,sizeof(pre));
    17     memset(iscut,0,sizeof(iscut));
    18     dfs_clock=0;
    19     for(int i=1;i<=N;i++)
    20         G[i].clear();
    21 }
    22 
    23 int tarjan(int u,int fa)
    24 {
    25     int lowu=pre[u]=++dfs_clock;
    26     int child=0;
    27     for(int i=0;i<G[u].size();i++)
    28     {
    29         int v=G[u][i];
    30         if(!pre[v])
    31         {
    32             child++;
    33             int lowv=tarjan(v,u);
    34             lowu=min(lowu,lowv);
    35             if(lowv>=pre[u])
    36                 iscut[u]=1;
    37         }
    38         else if(pre[v]<pre[u]&&v!=fa)
    39             lowu=min(lowu,pre[v]);
    40     }
    41     if(fa<0&&child==1)
    42         iscut[u]=0;
    43     low[u]=lowu;
    44     return lowu;
    45 }
    46 
    47 int main()
    48 {
    49     while(scanf("%d",&N),N)
    50     {
    51         init();
    52         int u;
    53         while(scanf("%d",&u),u)
    54         {
    55             int v;
    56             while(getchar()!='
    ')
    57             {
    58                 scanf("%d",&v);
    59                 G[u].push_back(v);
    60                 G[v].push_back(u);
    61             }
    62         }
    63         for(int i=1;i<N;i++)
    64         if(!pre[i])
    65             tarjan(i,-1);
    66         int cnt=0;
    67         for(int i=1;i<=N;i++)
    68             if(iscut[i])
    69             cnt++;
    70         printf("%d
    ",cnt);
    71     }
    72     return 0;
    73 }
  • 相关阅读:
    intellij idea for mac 2018 破解版
    Mac下Supervisor进程监控管理工具的安装与配置
    Mysql千万级大表优化策略
    php7实现基于openssl的加密解密方法
    openresty--centos7下开发环境安装
    webstorm下搭建编译less环境 以及设置压缩css
    七牛图片上传
    聊一聊PHP的依赖注入(DI) 和 控制反转(IoC)
    joomla! 3.X 开发系列教程
    JSON反序列化接口的问题
  • 原文地址:https://www.cnblogs.com/homura/p/4829412.html
Copyright © 2011-2022 走看看