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

題意：給你一個有向圖，問在圖中最少要加多少條邊能使得該圖變成一個強連通圖

思路：縮點模板題

Source Program

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<string>
#include<cstdlib>
#include<queue>
#include<set>
#include<map>
#include<stack>
#include<ctime>
#include<vector>
#define INF 0x3f3f3f3f
#define PI acos(-1.0)
#define N 20001
#define MOD 16007
#define E 1e-6
#define LL long long
using namespace std;
int n,m;
vector<int> G[N];
stack<int> S;
int dfn[N],low[N];
bool vis[N];
int sccno[N];
bool in[N],out[N];//記錄入度、出度是否為0
int block_cnt;
int sig;
void Tarjan(int x){
    vis[x]=true;
    dfn[x]=low[x]=++block_cnt;
    S.push(x);

    for(int i=0;i<G[x].size();i++){
        int y=G[x][i];
        if(vis[y]==false){
            Tarjan(y);
            low[x]=min(low[x],low[y]);
        }
        else if(!sccno[y])
            low[x]=min(low[x],dfn[y]);
    }

    if(dfn[x]==low[x]){
        sig++;
        while(true){
            int temp=S.top();
            S.pop();
            sccno[temp]=sig;
            if(temp==x)
                break;
        }
    }
}
void shrink(){//縮點
    memset(in,false,sizeof(in));
    memset(out,false,sizeof(out));
    for(int i=1;i<=sig;i++){
        in[i]=true;
        out[i]=true;
    }

    for(int x=0;x<n;x++){
        for(int i=0;i<G[x].size();i++){
            int y=G[x][i];
            if(sccno[x]!=sccno[y]){//統計每個點出度、入度是否為0
                out[sccno[x]]=false;
                in[sccno[y]]=false;
            }
        }
    }
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        for(int i=0;i<n;i++)
            G[i].clear();
        while(m--){
            int x,y;
            scanf("%d%d",&x,&y);
            x--;
            y--;
            G[x].push_back(y);
        }

        sig=0;
        block_cnt=0;
        memset(vis,false,sizeof(vis));
        memset(dfn,0,sizeof(dfn));
        memset(low,0,sizeof(low));
        memset(sccno,0,sizeof(sccno));

        for(int i=0;i<n;i++)
            if(vis[i]==false)
                Tarjan(i);

        shrink();
        int a=0,b=0;
        for(int i=1;i<=sig;i++){//統計入度、出度為0的點的個數
            if(in[i])
                a++;
            if(out[i])
                b++;
        }
        int res=max(a,b);
        if(sig==1)//強連通分量為1時
            res=0;
 
        printf("%d\n",res);
    }
}