zoukankan      html  css  js  c++  java
  • POJ 1236 Network of Schools(强连通缩点)

    Network of Schools

     

    Description

    A number of schools are connected to a computer network. Agreements have been developed among those schools: each school maintains a list of schools to which it distributes software (the “receiving schools”). Note that if B is in the distribution list of school A, then A does not necessarily appear in the list of school B 
    You are to write a program that computes the minimal number of schools that must receive a copy of the new software in order for the software to reach all schools in the network according to the agreement (Subtask A). As a further task, we want to ensure that by sending the copy of new software to an arbitrary school, this software will reach all schools in the network. To achieve this goal we may have to extend the lists of receivers by new members. Compute the minimal number of extensions that have to be made so that whatever school we send the new software to, it will reach all other schools (Subtask B). One extension means introducing one new member into the list of receivers of one school. 

    Input

    The first line contains an integer N: the number of schools in the network (2 <= N <= 100). The schools are identified by the first N positive integers. Each of the next N lines describes a list of receivers. The line i+1 contains the identifiers of the receivers of school i. Each list ends with a 0. An empty list contains a 0 alone in the line.

    Output

    Your program should write two lines to the standard output. The first line should contain one positive integer: the solution of subtask A. The second line should contain the solution of subtask B.

    Sample Input

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

    Sample Output

    1
    2

     1 #include<cstdio>
     2 #include<cstring>
     3 #include<vector>
     4 #include<stack>
     5 #include<algorithm>
     6 using namespace std;
     7 
     8 stack<int>s;
     9 vector<int>G[105];
    10 int dfn[105],sccno[105],in[105],out[105],lowlink[105];
    11 int n,dfs_clock,scc_cnt;
    12 
    13 void init()
    14 {
    15     memset(lowlink,0,sizeof(lowlink));
    16     memset(dfn,0,sizeof(dfn));
    17     memset(in,0,sizeof(in));
    18     memset(out,0,sizeof(out));
    19     memset(sccno,0,sizeof(sccno));
    20     for(int i=1;i<=n;i++)G[i].clear();
    21     dfs_clock=scc_cnt=0;
    22 }
    23 
    24 void tarjan(int u)
    25 {
    26     lowlink[u]=dfn[u]=++dfs_clock;
    27     s.push(u);
    28     for(int i=0;i<G[u].size();i++)
    29     {
    30         int v=G[u][i];
    31         if(!dfn[v])
    32         {
    33             tarjan(v);
    34             lowlink[u]=min(lowlink[u],lowlink[v]);
    35         }
    36         else if(!sccno[v])
    37             lowlink[u]=min(lowlink[u],dfn[v]);
    38     }
    39     if(lowlink[u]==dfn[u])
    40     {
    41         scc_cnt++;
    42         while(1)
    43         {
    44             int x=s.top();
    45             s.pop();
    46             sccno[x]=scc_cnt;
    47             if(x==u)break;
    48         }
    49     }
    50 }
    51 
    52 int main()
    53 {
    54     while(scanf("%d",&n)!=EOF)
    55     {
    56         init();
    57         for(int i=1;i<=n;i++)
    58         {
    59             int tmp;
    60             while(scanf("%d",&tmp),tmp)
    61                 G[i].push_back(tmp);
    62         }
    63         for(int i=1;i<=n;i++)
    64             if(!dfn[i])
    65                 tarjan(i);
    66         for(int i=1;i<=n;i++)
    67             for(int j=0;j<G[i].size();j++)
    68                 if(sccno[i]!=sccno[G[i][j]])
    69                 {
    70                     out[sccno[i]]++;
    71                     in[sccno[G[i][j]]]++;
    72                 }
    73         int cnt1=0,cnt2=0;
    74         for(int i=1;i<=scc_cnt;i++)
    75         {
    76             if(!in[i])
    77                 cnt1++;
    78             if(!out[i])
    79                 cnt2++;
    80         }
    81         if(scc_cnt==1)
    82             printf("1
    0
    ");
    83         else
    84             printf("%d
    %d
    ",cnt1,max(cnt1,cnt2));
    85     }
    86     return 0;
    87 }


  • 相关阅读:
    js 加密混淆工具
    postgresql 相关函数总结
    eclipse启动Tomcat时报错:严重: Exception loading sessions from persistent storage
    java sm4国密算法加密、解密
    java mail 封装工具类使用
    echarts 折线图百分比 tooltip 实例 两种方法
    开启人生的巅峰!!!
    判断是不是谷歌浏览器
    C#导出Excel的具体代码,供大家参考,具体内容请求URL
    SVN 安装后右键出现点击鼠标右键弹出错误提示:CrashHandler initialization error
  • 原文地址:https://www.cnblogs.com/homura/p/4853241.html
Copyright © 2011-2022 走看看