Proving Equivalences(强连通分量）

题意：已知有n个点，现在有m条边，问最少加几条边使其变为强连通图。

题解：就是求过强连通分量之后，然后找入度为0和出度为0的最大值。

 

Problem Description

Consider the following exercise, found in a generic linear algebra textbook.

Let A be an n × n matrix. Prove that the following statements are equivalent:

1. A is invertible.
2. Ax = b has exactly one solution for every n × 1 matrix b.
3. Ax = b is consistent for every n × 1 matrix b.
4. Ax = 0 has only the trivial solution x = 0. 

The typical way to solve such an exercise is to show a series of implications. For instance, one can proceed by showing that (a) implies (b), that (b) implies (c), that (c) implies (d), and finally that (d) implies (a). These four implications show that the four statements are equivalent.

Another way would be to show that (a) is equivalent to (b) (by proving that (a) implies (b) and that (b) implies (a)), that (b) is equivalent to (c), and that (c) is equivalent to (d). However, this way requires proving six implications, which is clearly a lot more work than just proving four implications!

I have been given some similar tasks, and have already started proving some implications. Now I wonder, how many more implications do I have to prove? Can you help me determine this?

 

 

Input

On the first line one positive number: the number of testcases, at most 100. After that per testcase:

* One line containing two integers n (1 ≤ n ≤ 20000) and m (0 ≤ m ≤ 50000): the number of statements and the number of implications that have already been proved.
* m lines with two integers s1 and s2 (1 ≤ s1, s2 ≤ n and s1 ≠ s2) each, indicating that it has been proved that statement s1 implies statement s2.

 

 

Output

Per testcase:

* One line with the minimum number of additional implications that need to be proved in order to prove that all statements are equivalent.

 

 

Sample Input

2 4 0 3 2 1 2 1 3

 

 

Sample Output

4 2

 

 

Source

NWERC 2008

#include<stdio.h>
#include<vector>
#include<string.h>
#include<algorithm>
using namespace std;
#define maxn 51010
#define N 21010
struct Edge{
	int from,to,nex;
	bool sign;
}edge[maxn<<1];
int head[maxn],edgenum;
int n,m;
void add(int u,int v)
{
	Edge E={u,v,head[u],false};
	edge[edgenum]=E;
	head[u]=edgenum++;
}
int DFN[N],Low[N],Stack[N],top,Time;
int taj;
int Belong[N];
bool Instack[N];
vector<int>bcc[N];
vector<int>G[N];
int du[N];
void init()
{
	memset(head,-1,sizeof(head));
	edgenum=0;
}
void tarjan(int u,int fa)
{
	DFN[u]=Low[u]=++Time;
	Stack[top++]=u;
	Instack[u]=1;
	for(int i=head[u];~i;i=edge[i].nex)
	{
		int v=edge[i].to;
		if(DFN[v]==-1)
		{
			tarjan(v,u);
			Low[u]=min(Low[u],Low[v]);
			if(DFN[u]<Low[v])
			{
				edge[i].sign=1;
			}
		}
		else if(Instack[v]) Low[u]=min(Low[u],DFN[v]);
	}
	if(Low[u]==DFN[u])
	{
		int now;
		taj++;
		bcc[taj].clear();
		do{
			now=Stack[--top];
			Instack[now]=0;
			Belong[now]=taj;
			bcc[taj].push_back(now); 
		}while(now!=u);
	}
}
void suodian()
{
	memset(du,0,sizeof(du));
	for(int i=1;i<=taj;i++) G[i].clear();
	for(int i=0;i<edgenum;i++)
	{
		int u=Belong[edge[i].from],v=Belong[edge[i].to];
		if(u!=v) G[u].push_back(v),du[v]++;
	}
	int w1=0,w2=0;
	if(taj==1)
	{
		printf("0\n");
		return;
	 } 
	for(int i=1;i<=taj;i++)
	{
		if(G[i].size()==0) w1++;
		if(du[i]==0) w2++;
	}
	printf("%d\n",w1>w2?w1:w2);
}
void tarjan_init(int all)
{
	memset(DFN,-1,sizeof(DFN));
	memset(Instack,0,sizeof(Instack));
	top=Time=taj=0;
	for(int i=1;i<=all;i++)
	{
		if(DFN[i]==-1)
		{
			tarjan(i,i);
		}
	}
}
int main()
{
	int T;
	scanf("%d",&T);
	while(T--)
	{
		scanf("%d%d",&n,&m);
	    init();
	  for(int i=1;i<=m;i++)
	 {
	    	int x,y;
		 scanf("%d%d",&x,&y);
		  add(x,y);
	 }
	 tarjan_init(n);
	 suodian();
   }
   return 0;
}