[tarjan] poj 1236 Network of Schools

主题链接：

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

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

Sample Output

1
2

Source

field=source&key=IOI+1996" style="text-decoration:none">IOI 1996

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题目意思：

给出每一个学校可以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，才干是全部学校都收到信息。

求Subtask B：求至少须要加入几个接受信息学校，使得当信息发送给随意一个学校时，都能传送到其它学校。

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）。

把每一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;
}

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