POJ 1236 Network of Schools 强连通图

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

图论题目，须要解决这个问题：

1 使用Tarjan算法求子强连通图

2 标识顶点属于哪个子强连通图

3 计算各个子强连通图的零入度数和零出度数

图论中高级内容了，是有点难度的，不细心一点肯定会出错的。

这次本博主认真注解好差点儿每一个语句。希望大家能够follow我的程序。

#include <cstdio>
#include <stack>
#include <cstring>
#include <algorithm>
#include <vector>
using namespace std;

const int MAX_N = 101;//最大的顶点数
vector<int> graAdj[MAX_N];//vector表示邻接表法
int visNo[MAX_N];//记录深搜各个顶点的訪问顺序标号
int lowLink[MAX_N];//连通图的最低标识号，记录好是否递归到已经訪问过的顶点了。假设是，那么就以最低的顶点訪问顺序标号为，这样能够统一子连通图的标号。

通过推断当前最低连通图的标识号和訪问顺序号是否一致来推断是否找到了一个子强连通图
int dfsNo;//记录深搜总的訪问号
int connectNo;//当前的子强连通图的标号，终于为全部子强连通图的数量
int markNo[MAX_N];//markNo[v]代表顶点v属于子强连通图markNo[v]，其值就为子强连通图
int in[MAX_N], out[MAX_N];//分别记录一个子强连通图的入度数和出度数
stack<int> stk;//深搜顶点入栈，找到子强连通图时候出栈，直到当前顶点数，全部点都属于同一个子强连通图

//深搜查找子强连通图，并记录好顶点属于哪个子强连通图
void getStrongConnected(int u)
{
	visNo[u] = lowLink[u] = ++dfsNo;//第一次进入当前顶点的时候的初值
	int n = (int)graAdj[u].size();
	stk.push(u);
	for (int i = 0; i < n; i++)//遍历当前顶点的全部连接点
	{
		int v = graAdj[u][i];
		if (!visNo[v])//没有訪问过的时候
		{
			getStrongConnected(v);//递归
			lowLink[u] = min(lowLink[u], lowLink[v]);//记录最低序号
		}
		//已经訪问过，可是还在栈里面，即还没有记录该顶点属于哪个强连通图
		else if (!markNo[v]) lowLink[u] = min(lowLink[u], lowLink[v]);
	}
	if (visNo[u] == lowLink[u])//当前訪问顺序号等于最低标号，
	{//那么就是找到了一个子强连通图
		++connectNo;//每次要添加全局的连通标号
		int v;
		do
		{
			v = stk.top(); stk.pop();
			markNo[v] = connectNo;//顶点对用强连通图号
		} while (u != v);
	}
}

void Tarjan(int n)
{
	//前期清零工作
	dfsNo = 0, connectNo = 0;
	fill(visNo, visNo+n+1, 0);
	fill(lowLink, lowLink+n+1, 0);
	fill(markNo, markNo+n+1, 0);
	while (!stk.empty()) stk.pop();

	for (int u = 1; u <= n; u++)
	{
		//某些顶点或许是分离的。就是图的顶点有不相连的，故此要遍历全部顶点
		if (!visNo[u]) getStrongConnected(u);
	}
}

int main()
{
	int N, u, v;
	scanf("%d", &N);
	for (u = 1; u <= N; u++)
	{
		scanf("%d", &v);
		while (v)//为零表示结束
		{
			graAdj[u].push_back(v);//使用vector建立一个邻接表
			scanf("%d", &v);
		}
	}
	Tarjan(N);//计算子强连通图的个数，并表出各个顶点属于哪个子强连通图

	for (u = 1; u <= N; u++)
	{//遍历全部顶点，然后遍历顶点的邻接边。相当于遍历全部边
		for (int i = 0; i < (int)graAdj[u].size(); i++)
		{
			int v = graAdj[u][i];
			if (markNo[u] != markNo[v])//不是属于同一个子强连通图
			{//分别添加该强连通图的入度和出度
				out[markNo[u]]++;
				in[markNo[v]]++;
			}
		}
	}

	int zeroIn = 0, zeroOut = 0;
	for (int i = 1; i <= connectNo; i++)
	{
		if (in[i] == 0) zeroIn++;
		if (out[i] == 0) zeroOut++;
	}
	//入度为零则须要放置一个软件拷贝
	printf("%d\n", zeroIn);
	//变为一个强连通图，分2个情况：1 本身是一个强连通图；2 零入度或出度最大值
	printf("%d\n", connectNo == 1? 0 : max(zeroIn, zeroOut));

	return 0;
}