Uva 10608- 并查集，最大的朋友个数

Problem I

FRIENDS

 

There is a town with N citizens. It is known that some pairs of people are friends. According to the famous saying that “The friends of my friends are my friends, too” it follows that if A and B are friends and B and C are friends then A and C are friends, too.

 

Your task is to count how many people there are in the largest group of friends.

 

Input

Input consists of several datasets. The first line of the input consists of a line with the number of test cases to follow. The first line of each dataset contains tho numbers N and M, where N is the number of town's citizens (1≤N≤30000) and M is the number of pairs of people (0≤M≤500000), which are known to be friends. Each of the following M lines consists of two integers A and B (1≤A≤N, 1≤B≤N, A≠B) which describe that A and B are friends. There could be repetitions among the given pairs.

 

Output

The output for each test case should contain one number denoting how many people there are in the largest group of friends.

 

Sample Input	Sample Output
2 3 2 1 2 2 3 10 12 1 2 3 1 3 4 5 4 3 5 4 6 5 2 2 1 7 10 1 2 9 10 8 9	3 6

之前写的并查集都是求联通子集的个数，这个题目是求出最大的子集的个数还是模板题

#include <cstdio>
#include <iostream>
#include <cmath>

using namespace std;

int parent[30005];
int num[30005];	
int n, m;
int find (int x) {
	return parent[x] == x ? x : find(parent[x]);
}
void init () {
	int i;
	for (i = 0; i < n; i ++)
		parent[i] = i;
	for (i = 0; i < n; i ++)
		num[i] = 0;
}
void Union(int a, int b) {
	int fa = find(a);
	int fb = find(b);
	if(fa != fb) {
		parent[fa] = fb;
	}
}
int main () {
	int cas;
	int a, b;
	int i;
	scanf("%d", &cas);
	while (cas --) {
		scanf("%d%d", &n, &m);
		init ();
		for (i = 0; i < m; i ++) {
			scanf("%d%d", &a, &b);
			Union(a, b);
		}
		for (i = 0; i < n; i ++) {
			int x = find(i);
			num[x] ++;
		}
		int max = 0;
		for (i = 0; i < n; i ++) {
			if(num[i] > max)
				max = num[i];
		}
		printf("%d\n", max);
	}
	return 0;
}