註意 receive sizeof date() pos describes nim distrib uci

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

1，一個有向圖，問最少選擇幾個點做源頭，使得可以通過它們到達所以點（一個點可以到達任意多個後序點）------求入度為0的點數。

2，一個有向圖，問最少加幾條邊後強連通------求max(入度為0的點數，出度為0的點數)，之前證明過。但是需要註意已經強連通的情況下會出錯（需要特判）。

3，一個有向圖，問最少選擇幾個點做源頭，使得可以通過它們到達所以點（一個點可以到達一個後序點）------縮點後用最小路徑覆蓋（二分圖匹配）

此題只求1，2兩個問，一定要註意特判。

//有向圖縮點 ,註意scc_cnt=1時。 
#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;
}

POJ1236 Network of Schools (強連通分量，註意邊界)