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:
    java 正则表达式匹配指定变量并替换
    Tomcat 架构原理解析到架构设计借鉴
    优雅的缓存写法,以及synchronized 和 ReentrantLock性能 PK
    应用开发笔记|MYD-YA157-V2开发板CAN BUS 总线通信实例
    Arm Keil MDK V5.33版本更新,欢迎下载!
    Arm Development Studio 2020.1版本下载更新
    设计模式 | 享元模式(Flyweight)
    设计模式 | 中介者模式/调停者模式(Mediator)
    设计模式 | 职责链模式(Chain of responsibility)
    设计模式 | 命令模式(Command)
  • 原文地址:https://www.cnblogs.com/homura/p/4853241.html
Copyright © 2011-2022 走看看