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?

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.

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.

2
4 0
3 2
1 2
1 3

4
2

题意:给你n个关系，m条推导，问你最少需要添加几条推导才能使任意两个关系之间可以互推。

然后这个问题便转化成一幅图，需要添加几条边，使得这个图变成强连通图。

思路：首先用tarjan把图变成DAG图，然后统计没有被连上的点（入度出度为0）。找最大的入度和出度，即代表最少需要添加几条边。

#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <stack>
using namespace std;
#define MAXN 200010
#define clr(x,k) memset((x),(k),sizeof(x))
struct node
{
    int st,to,next;
}
edge[MAXN];
int n,m,ct,id;
int head[MAXN],low[MAXN],dfn[MAXN],belong[MAXN],in[MAXN],to[MAXN];
//DFN[i]表示 遍历到 i 点时是第几次dfs
//Low[u] 表示 以u点为父节点的 子树 能连接到 [栈中] 最上端的点 的DFN值
//belong存的是缩点后的点。
//ct 是指相互的点到达可以缩成一个点的个数。
bool instack[MAXN];
stack<int>q;
void add_e(int i,int u,int v)
{
    edge[i].st=u;
    edge[i].to=v;
    edge[i].next=head[u];
    head[u]=i;
}
void tarjan(int i)
{
    int j;
    dfn[i]=low[i]=++id;
    q.push(i);
    instack[i]=1;
    for(int u=head[i]; ~u; u=edge[u].next)  //求强连通分量
    {
        j=edge[u].to;
        if(dfn[j]==0)
        {
            tarjan(j);
            if(low[i]>low[j])
                low[i]=low[j];
        }
        else if(instack[j]&&low[i]>low[j])
            low[i]=dfn[j]; 
    }
    if(dfn[i]==low[i])  //缩点
    {
        ct++;
        do
        {
            j=q.top();
            q.pop();
            instack[j]=0;
            belong[j]=ct;
        }
        while(i!=j);
    }
}
int main()
{
    int t,i,u,v,sum1,sum2;
    cin>>t;
    while(t--)
    {
        clr(head,-1);
        clr(low,0);
        clr(dfn,0);
        clr(belong,0);
        clr(in,0);
        clr(to,0);
        while(!q.empty())
            q.pop();
        cin>>n>>m;
        for(i=0; i<m; i++)
        {
            cin>>u>>v;
            add_e(i,u,v);
        }
        id=ct=0;
        for(i=1; i<=n; i++)
        {
            if(!dfn[i])
                tarjan(i);
        }
        if(ct==1) //所有的点可以任意到达，原图即为强连通图。
        {
            cout<<0<<endl;
            continue;
        }
        for(i=1; i<=ct; i++)
        {
            in[i]=to[i]=0;
        }
        for(i=0; i<m; i++)
        {
            if(belong[edge[i].st]!=belong[edge[i].to])
            {
                in[belong[edge[i].st]]++;
                to[belong[edge[i].to]]++;
            }
        }
        sum1=sum2=0;
        for(i=1; i<=ct; i++)
        {
            if(in[i]==0)
                sum1++;
            if(to[i]==0)
                sum2++;
        }
        cout<<max(sum1,sum2)<<endl;
    }
    return 0;
}