这个结合上题，类似的代码解决的：http://www.cnblogs.com/kuangbin/archive/2011/08/07/2130062.html

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

Source

IOI 1996   题目链接：
http://poj.org/problem?id=1236  

强连通分量缩点求入度为0的个数和出度为0的分量个数

题目大意：N(2<N<100)各学校之间有单向的网络，每个学校得到一套软件后，可以通过单向网络向周边的学校传输，问题1：初始至少需要向多少个学校发放软件，使得网络内所有的学校最终都能得到软件。2，至少需要添加几条传输线路(边)，使任意向一个学校发放软件后，经过若干次传送，网络内所有的学校最终都能得到软件。

也就是：

—        给定一个有向图，求：

1) 至少要选几个顶点，才能做到从这些顶点出发，可以到达全部顶点

2) 至少要加多少条边，才能使得从任何一个顶点出发，都能到达全部顶点

—        顶点数<= 100

解题思路：

—        1. 求出所有强连通分量

—        2. 每个强连通分量缩成一点，则形成一个有向无环图DAG。

—        3. DAG上面有多少个入度为0的顶点，问题1的答案就是多少

在DAG上要加几条边，才能使得DAG变成强连通的，问题2的答案就是多少

加边的方法：

要为每个入度为0的点添加入边，为每个出度为0的点添加出边

假定有 n 个入度为0的点，m个出度为0的点，如何加边？

把所有入度为0的点编号 0,1,2,3,4 ….N -1

每次为一个编号为i的入度0点可达的出度0点，添加一条出边，连到编号为(i+1)%N 的那个出度0点,

这需要加n条边

若 m <= n，则

加了这n条边后，已经没有入度0点，则问题解决，一共加了n条边

若 m > n，则还有m-n个入度0点，则从这些点以外任取一点，和这些点都连上边，即可，这还需加m-n条边。

所以，max(m,n)就是第二个问题的解

此外：当只有一个强连通分支的时候，就是缩点后只有一个点，虽然入度出度为0的都有一个，但是实际上不需要增加清单的项了，所以答案是1，0；

程序：

/*
POJ 1236 

*/ 
#include<stdio.h>
#include<string.h>
#include<iostream>
using namespace std;
#define MAXN 110
struct Node
{
    int to,next;
}edge1[MAXN*100],edge2[MAXN*100];
int head1[MAXN];
int head2[MAXN];
int mark1[MAXN],mark2[MAXN];
int tot1,tot2;
int cnt1,cnt2,st[MAXN],belong[MAXN];
int num,setNum[MAXN];
struct Edge
{
    int l,r;
}e[MAXN*100];
void add(int a,int b)
{
    edge1[tot1].to=b;edge1[tot1].next=head1[a];head1[a]=tot1++;
    edge2[tot2].to=a;edge2[tot2].next=head2[b];head2[b]=tot2++;
}    
void DFS1(int x)//搜索原图
{
    mark1[x]=1;
    for(int i=head1[x];i!=-1;i=edge1[i].next)
        if(!mark1[edge1[i].to])  DFS1(edge1[i].to);
    st[cnt1++]=x;
}   
void DFS2(int x)
{
    mark2[x]=1;
    num++;
    belong[x]=cnt2;
    for(int i=head2[x];i!=-1;i=edge2[i].next)
       if(!mark2[edge2[i].to])  DFS2(edge2[i].to);
}  
int main()
{
    int n;
    int m;//边数 
    while(scanf("%d",&n)!=EOF)
    {
        int w,v;
        tot1=tot2=1;
        for(int i=1;i<=n;i++)
        {
            head1[i]=head2[i]=-1;
            mark1[i]=mark2[i]=0;
        }   
        m=0;
        for(w=1;w<=n;w++)
        {
            while(scanf("%d",&v),v)
            {
                m++;
                e[m].l=w;e[m].r=v;
                add(w,v);
            }    
        }
        cnt1=cnt2=1;
        for(int i=1;i<=n;i++)
        {
            if(!mark1[i])DFS1(i);
        }     
        for(int i=cnt1-1;i>=1;i--)
        {
            if(!mark2[st[i]])
            {
                num=0;
                DFS2(st[i]);
                setNum[cnt2++]=num;
            }    
        } 
        int out[MAXN],in[MAXN];//计算出度和入度   
        memset(out,0,sizeof(out));
        memset(in,0,sizeof(in));
        for(int i=1;i<=m;i++)
        {
            if(belong[e[i].l]!=belong[e[i].r])
            {
                out[belong[e[i].l]]++;
                in[belong[e[i].r]]++;
            }    
        }
        int res1=0;//出度为0的个数
        int res2=0;//入度为0的个数
        for(int i=1;i<cnt2;i++)
        {
            if(!out[i]) res1++;
            if(!in[i])  res2++;
        }
        //下面这个是关键，整张图就一个连通分量时，输出0，1.WR了好几次呢 
        if(cnt2==2) {printf("1\n0\n");continue;}
        printf("%d\n",res2);
        printf("%d\n",res1>res2?res1:res2);            
    }  
    return 0;  
}