http://acm.nyist.net/JudgeOnline/problem.php?pid=120&&强连通分量

题意：让一个图变成强连通图，最少需要添加多少边

思路：先求出强连通图的个数，然后缩边后比较图中入度为0和出度为0的顶点个数

 
#include<iostream>
#include<string>
#include<string.h>
#include<algorithm>
#define N 101
#define M 10010
using namespace std;
bool istack[N];
int head[N],dfn[N],low[N],stack[N],belong[N],out[N],in[N];
int n,m,res,indexx,num,top;
typedef struct str
{
	int to;
	int next;
}Node;
Node node[M];
void initt()
{  memset(head,-1,sizeof(head));
   memset(dfn,0,sizeof(dfn));
   memset(low,0,sizeof(low));
   memset(out,0,sizeof(out));
   memset(belong ,0,sizeof(belong));
   memset(istack,false,sizeof(istack));
   memset(stack,0,sizeof(stack));
   memset(in,0,sizeof(in));
	res=top=num=0;
	indexx=1;
}
void dfs(int i)
{
	dfn[i]=low[i]=indexx++;
	istack[i]=true;
	stack[top++]=i;
	for(int j=head[i];j!=-1;j=node[j].next)
	{
		 int v=node[j].to;
		 if(!dfn[v])
		 {
			 dfs(v);
			 low[i]=min(low[i],low[v]);
		 }
		 else if(istack[v]) low[i]=min(dfn[v],low[i]);
	}
	int u;
	if(dfn[i]==low[i])
	{
		res++;
		do
		{ 
			u=stack[--top];
			belong[u]=res;
			istack[u]=false;
		}while(i!=u);
	}
}
void tarjan()
{
	for(int i=1;i<=n;++i)
		if(!dfn[i]) dfs(i);
}
void solve()
{
	for(int i=1;i<=n;++i)
	  for(int j=head[i];j!=-1;j=node[j].next)
			if(belong[i]!=belong[node[j].to])
				{
					out[belong[i]]++;
					in[belong[node[j].to]]++;
			     }
			if(res==1) cout<<"0"<<endl;//当图本身就是强连通图时
			else
			{
			int p=0,q=0;
			for(int i=1;i<=res;++i)
			{ if(!out[i]) p++;
			   if(!in[i]) q++;
			}
			cout<<max(p,q)<<endl;
			}
}
int main()
{  //freopen("c:\\Input.txt","r",stdin);
   //freopen("c:\\Out.txt","w",stdout);
	int T;
	cin>>T;
	while(T--)
	{    initt();
		cin>>n;
		for(int i=1;i<=n;++i)
		{   int flag=0;
			int a;
		     while(cin>>a&&a)
			 {  node[num].to=a;
				node[num].next=head[i];
				head[i]=num++;
			 }
		 }
		tarjan();
		solve();
     }return 0;
}