POJ 1236 Network of Schools（强连通 + 想法）- from lanshui_Yang

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

    题目大意：有n个学校，学校之间可以传递信息，例如学校 a 可以 传达信息给 b ， 即a ——> b , 但 b 不一定 能传递信息给 a 。 告诉你每个学校能够向哪些学校传递信息，然后有两个问题：

    问题一：至少要向几个学校传递 原始 信息，才能保证所有学校都能收到信息。

    问题二：至少要添加多少组关系（每组关系类型如右：a 可以 向 b 传递信息），才能保证 给任意一个学校原始信息后，其他所有学校都能收到信息。

    解题思路：这道题其实就是一个有n个顶点的有向图，先用 tarjan 缩点 ， 然后分别统计出 入度为0 和 出度为0 的强连通分量个数ansA 和 ansB，那么， 问题一的答案就是ansA , 问题二的答案就是max（ansA ， ansB），但是有特例：当只有一个强连通分量时，问题二的答案就是0 。

    下面用两张图来说明一下问题二的连边方式：

    请看代码：

#include<iostream>
#include<cstring>
#include<string>
#include<algorithm>
#include<cmath>
#include<cstdio>
#include<vector>
#define mem(a , b) memset(a , b , sizeof(a))
using namespace std ;
const int MAXN = 105 ;
vector<int> G[MAXN] ;
int id[MAXN] ; // 记录缩点后每个顶点所属的强连通分量
int tmpdfn ;
int low[MAXN] ;
int dfn[MAXN] ;
int stap[MAXN] ; // 模拟栈
bool vis[MAXN] ;
int ind[MAXN] ; // 统计每个强连通分量的入度
int od[MAXN] ;  // 统计每个强连通分量的出度
int top ;
int component ; // 统计强连通分量数量
int n ;
int ansA ;
int ansB ;
const int INF = 0x7fffffff ;
void chu()
{
    int i ;
    for(i = 0 ; i <= n ; i ++)
        G[i].clear() ;
    mem(id , -1) ;
    mem(dfn , 0) ;
    mem(low , 0) ;
    mem(vis , 0) ;
    mem(od , 0) ;
    mem(ind , 0) ;
    component = 0 ;
    tmpdfn = 0 ;
    top = - 1 ;
}
void init()
{
    chu() ;
    int i ;
    for(i = 1 ; i <= n ; i ++)  // 建图
    {
        int b ;
        for(;;)
        {
            scanf("%d" , &b) ;
            if(b == 0)
                break ;
            G[i].push_back(b) ;
        }
    }
}
void tarjan(int u)
{
    vis[u] = true ;
    stap[++ top] = u ;
    dfn[u] = low[u] = ++ tmpdfn ;
    int i ;
    for(i = 0 ; i < G[u].size() ; i ++)
    {
        int v = G[u][i] ;
        if(!vis[v])
        {
            tarjan(v) ;
            low[u] = min(low[u] , low[v]) ;
        }
        else
        {
            low[u] = min(dfn[v] , low[u]) ;
        }
    }
    if(low[u] == dfn[u])  // 缩点
    {
        ++ component ;
        int tmp ;
        do
        {
            tmp = stap[top --] ;
            dfn[tmp] = INF ;  // 勿忘此处！！,这里是排除其他连通快的影响 ！！
            id[tmp] = component ;
        }
        while(tmp != u) ;
    }
}
void solve()
{
    ansA = ansB = 0 ;
    bool flag = false ;
    int i ;
    for(i = 1 ; i <= n ; i ++)
    {
        if(!vis[i])  // 注意此处，可能不止一个连通块
        {
            tarjan(i) ;
        }
    }
    int j ;
    for(i = 1 ; i <= n ; i ++)
    {
        for(j = 0 ; j < G[i].size() ; j ++)
        {
            int ta , tb ;
            ta = id[i] ;
            tb = id[ G[i][j] ] ;
            if(ta != tb)
            {
                ind[tb] ++ ;
                od[ta] ++ ;
            }
        }
    }
    for(i = 1 ; i <= component ; i ++)
    {
        if(ind[i] == 0)
            ansA ++ ;
        if(od[i] == 0)
            ansB ++ ;
    }
    if(component == 1)
    {
        ansB = 0 ;
    }
    else
    {
        ansB = max(ansB , ansA) ;
    }
    printf("%d\n%d\n" , ansA , ansB) ;
}
int main()
{
    while (scanf("%d" , &n) != EOF)
    {
        init() ;
        solve() ;
    }
    return 0 ;
}