POJ1236 Network of Schools (强连通分量，注意边界)

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

 

1，一个有向图，问最少选择几个点做源头，使得可以通过它们到达所以点（一个点可以到达任意多个后序点）------求入度为0的点数。

2，一个有向图，问最少加几条边后强连通------求max(入度为0的点数，出度为0的点数)，之前证明过。但是需要注意已经强连通的情况下会出错（需要特判）。

3，一个有向图，问最少选择几个点做源头，使得可以通过它们到达所以点（一个点可以到达一个后序点）------缩点后用最小路径覆盖（二分图匹配）

 

此题只求1，2两个问，一定要注意特判。

//有向图缩点 ,注意scc_cnt=1时。 
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn=210;
int Laxt[maxn],Next[maxn*200],To[maxn*200],cnt,n;
int dfn[maxn],low[maxn],times,scc_cnt,scc[maxn];
int instc[maxn],stc[maxn],top,ans1,ans2;
int ind[maxn],oud[maxn];
void update()
{
    cnt=times=scc_cnt=top=ans1=ans2=0;
    memset(Laxt,0,sizeof(Laxt));
    memset(dfn,0,sizeof(dfn));
    memset(low,0,sizeof(low));
    memset(scc,0,sizeof(scc));
    memset(instc,0,sizeof(instc));
    memset(stc,0,sizeof(stc));
    memset(ind,0,sizeof(ind));
    memset(oud,0,sizeof(oud));
}
void add(int u,int v)
{
    Next[++cnt]=Laxt[u];
    Laxt[u]=cnt;
    To[cnt]=v;
}
void dfs(int u)
{
    dfn[u]=low[u]=++times;
    stc[++top]=u; instc[u]=1;
    for(int i=Laxt[u];i;i=Next[i]){
        int v=To[i];
        if(!dfn[v]){
            dfs(v);
            low[u]=min(low[u],low[v]);
        }
        else if(instc[v]){
            low[u]=min(low[u],dfn[v]);
        }
    }
    if(dfn[u]==low[u]){
        scc_cnt++;
        while(true){
            int x=stc[top--];
            scc[x]=scc_cnt;
            instc[x]=0;
            if(x==u) break;
        }
    }
}
void tarjan()
{
    for(int i=1;i<=n;i++) 
        if(!dfn[i]) dfs(i);
    for(int i=1;i<=n;i++)
      for(int j=Laxt[i];j;j=Next[j]){
        if(scc[i]!=scc[To[j]]) {
            ind[scc[To[j]]]++;
            oud[scc[i]]++;
        }
    }
        
    for(int i=1;i<=scc_cnt;i++){
        if(ind[i]==0) ans1++;
        if(oud[i]==0) ans2++;
    }
}
int main()
{
    while(~scanf("%d",&n)){
        update();
        for(int i=1;i<=n;i++) {
            int x; while(scanf("%d",&x)){
                if(x==0) break;
                add(i,x);
            }
        }
        tarjan();
        if(scc_cnt==1) printf("1\n0\n");
        else printf("%d\n%d\n",ans1,ans2);
    } return 0;
}