[tarjan] poj 1236 Network of Schools

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本文探讨了如何通过最小化软件分发点的数量来确保网络中所有学校都能接收并传播新软件，包括两个子任务：计算直接接收源信息的最少学校数量（Subtask A），以及通过增加最少的接收者数量来确保任意学校都能传播到所有学校（Subtask B）。通过求解有向图的强连通分量和进行节点缩合，实现了对于网络结构的分析和优化。

题目链接：

http://poj.org/problem?id=1236

Network of Schools

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 11433		Accepted: 4551

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

Sample Output

1
2

Source

IOI 1996

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题目意思：

给出每个学校能够传送信息的学校名单。

求Subtask A:至少要给多少个学校发送源信息，才能是所有学校都收到信息。求Subtask B：求至少需要添加几个接受信息学校，使得当信息发送给任意一个学校时，都能传送到其他学校。

解题思路：

先求有向图的强连通分量。然后缩点。统计各点出度和入度。

Subtask A 就是求入度为0的点的个数。

Subtask B就是求max(入度为0个数，出度为零个数）。

把每个出度为零的节点连到下一颗树的入度为0的节点上。

代码：

//#include<CSpreadSheet.h>

#include<iostream>
#include<cmath>
#include<cstdio>
#include<sstream>
#include<cstdlib>
#include<string>
#include<string.h>
#include<cstring>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<stack>
#include<list>
#include<queue>
#include<ctime>
#include<bitset>
#include<cmath>
#define eps 1e-6
#define INF 0x3f3f3f3f
#define PI acos(-1.0)
#define ll __int64
#define LL long long
#define lson l,m,(rt<<1)
#define rson m+1,r,(rt<<1)|1
#define M 1000000007
//#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;

#define Maxn 110
int low[Maxn],dfn[Maxn],dindex,n;
int sta[Maxn],belong[Maxn],bcnt,ss;
bool iss[Maxn];
int de1[Maxn],de2[Maxn];
vector<vector<int> >myv;

void tarjan(int cur)
{
    //printf(":%d\n",cur);
    //system("pause");
    int ne;

    dfn[cur]=low[cur]=++dindex;
    iss[cur]=true;
    sta[++ss]=cur;

    for(int i=0;i<myv[cur].size();i++)
    {
        ne=myv[cur][i];
        if(!dfn[ne])
        {
            tarjan(ne);
            low[cur]=min(low[cur],low[ne]);
        }
        else if(iss[ne]&&dfn[ne]<low[cur])
            low[cur]=dfn[ne];
    }

    if(dfn[cur]==low[cur])
    {
        bcnt++;
        do
        {
            ne=sta[ss--];
            iss[ne]=false;
            belong[ne]=bcnt;
        }while(ne!=cur);
    }
}

void solve()
{
    int i;
    ss=bcnt=dindex=0;
    memset(dfn,0,sizeof(dfn));
    memset(iss,false,sizeof(iss));
    for(int i=1;i<=n;i++)
        if(!dfn[i])
            tarjan(i);
}

int main()
{
    //freopen("in.txt","r",stdin);
   //freopen("out.txt","w",stdout);
   while(~scanf("%d",&n))
   {
       myv.clear();
       myv.resize(n+1);
       for(int i=1;i<=n;i++)
       {
           int a;
           while(scanf("%d",&a)&&a)
               myv[i].push_back(a);
       }
       solve();

       if(bcnt==1) //本来就是一个强连通分量
       {
           printf("1\n0\n");
           continue;
       }
       int ansa=0,ansb=0;

       memset(de1,0,sizeof(de1));
       memset(de2,0,sizeof(de2));

       for(int i=1;i<=n;i++)
       {
           for(int j=0;j<myv[i].size();j++)
           {
               int ne=myv[i][j];
               if(belong[i]!=belong[ne])
               {
                   de1[belong[ne]]++;//入度
                   de2[belong[i]]++; //出度
               }

           }
       }
       for(int i=1;i<=bcnt;i++)
       {
           if(!de1[i])
                ansa++;
           if(!de2[i])
                ansb++;
       }
       ansb=max(ansb,ansa);

       printf("%d\n%d\n",ansa,ansb);
   }
    return 0;
}