toj-2469-朋友的朋友是朋友

There are some people traveling together. Some of them are friends. The friend relation is transitive, that is, if A and B are friends, B and C are friends, then A and C will become friends too.

These people are planning to book some rooms in the hotel. But every one of them doesn't want to live with strangers, that is, if A and D are not friends, they can't live in the same room.

Given the information about these people, can you determine how many rooms they have to book at least? You can assume that the rooms are large enough.

Input

The first line of the input is the number of test cases, and then some test cases followed.

The first line of each test case contain two integers N and M, indicating the number of people and the number of the relationship between them. Each line of the following M lines contain two numbers A and B (1 ≤ A ≤ N , 1 ≤ B ≤ N , A ≠ B), indicating that A and B are friends.

You can assume 1 ≤ N ≤ 100, 0 ≤ M ≤ N * (N-1) / 2. All the people are numbered from 1 to N.

Output

Output one line for each test case, indicating the minimum number of rooms they have to book.

Sample Input

35 31 22 34 55 41 22 33 44 510 0

Sample Output

2110

Hint

In the first sample test case, there are 5 people. We see that 1,2,3 can live in a room, while 4,5 can live in another room. So the answer should be 2. Please note even though 1 and 3 are not "direct" friends, but they are both 2's friends, so 1 and 3 are friends too.

In the second sample test case, there are 5 people, any two of them will become friends. So they can live in one room.

In the third sample test case, there are 10 people and no friend relationship between them. No two people can live the same room. They have to book 10 rooms.

方法一：将有联系的人并为一组，每组推举出一个老大，然后寻找有多少个老大，就有多少组

#include <iostream>
using namespace std;

int father[1000];
int find(int x)
{
    int r=x;
    while(father[r]!=r)
    {
        r = father[r];
    }
    int i=x, j;
    while(i!=r)
    {
        j=father[i];
        father[i]=r;
        i=j;
    }
    return r;
}
void join(int x, int y)
{
    int fx = find(x), fy = find(y);
    if(fx != fy)
        father[fx] = fy;
}
int main()
{
    int Case, N, M, a, b;
    cin >> Case;
    while(Case--)
    {
        cin >> N >> M;
        for(int i=1; i<=N; i++)
            father[i] = i;
        while(M--)
        {
            cin >> a >> b;
            join(a,b);
        }
        int count = 0;
        for(int i=1; i<=N; i++)
        {
            if(father[i] == i)  count++;
        }
        cout << count << endl;
    }
    return 0;
}