アルゴリズムノート練習9.6そして問題C:How Many Tableを調べる

アルゴリズムノート練習問題解合集
本題リンク
タイトル
タイトルはToday is Ignatius’birthday.He invites a lot of friends. Now it’s dinner time. Ignatius wants to know how many tables he needs at least. You have to notice that not all the friends know each other, and all the friends do not want to stay with strangers.
One important rule for this problem is that if I tell you A knows B, and B knows C, that means A, B, C know each other, so they can stay in one table.
For example: If I tell you A knows B, B knows C, and D knows E, so A, B, C can stay in one table, and D, E have to stay in the other one. So Ignatius needs 2 tables at least.
入力The input starts with an integer T(1<=T<=25)which indicate the number of test cases.Then T test cases follow. Each test case starts with two integers N and M(1<=N,M<=1000). N indicates the number of friends, the friends are marked from 1 to N. Then M lines follow. Each line consists of two integers A and B(A!=B), that means friend A and friend B know each other. There will be a blank line between two cases.
出力For each test case,just output how many tables Ignatius needs at least.Do NOT print any blanks.
サンプル入力

2
6 4
1 2
2 3
3 4
1 4

8 10
1 2
2 3
5 6
7 5
4 6
3 6
6 7
2 5
2 4
4 3

サンプル出力

3
2

構想
構想はこの問題を見ます:問題B:スムーズな工事.2つの問題は全く同じで、ただ皮を変えただけです~
コード#コード#

#include 
#include 
#include 
using namespace std;

vector<int> father;
set<int> roots;

int findFather(int x) {
	int f = x, temp;
	while (f != father[f])
		f = father[f];
	while (x != father[x]) {
		temp = x;
		x = father[x];
		father[temp] = f; 
	} 
	return f;
} 
void Union(int a, int b) {
	int fa = findFather(a), fb = findFather(b);
	if (fa != fb)
		father[fa] = fb; 
} 

int main() {
	int T, n, m, a, b;
	while (scanf("%d", &T) != EOF) {
		while (T--) { 
			scanf("%d%d", &n, &m);
			father.resize(n + 1);
			roots.clear();
			for (int i = 1; i <= n; ++i)
				father[i] = i;
			for (int i = 0; i < m; ++i) {
				scanf("%d%d", &a, &b);
				Union(a, b); 
			}
			for (int i = 1; i <= n; ++i)
				roots.insert(findFather(i));
			printf("%d
", roots.size());
		}
	} 
	return 0;
}