UVA 10608 Friends

Original question:
There is a town with N citizens. It is known this some pairs of people is friends. According to the famous saying that "The Friends of my friends is my friends, too" it follows that if A and B is friends and B and C is friends then A and C is friends, too.
Your task is to count how many people there be 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 was the number of town ' s citizens (1≤n≤30000) and M is the number of pairs of people (0≤m≤500000), which was known to be friends. Each of the following M lines consists of both integers A and B (1≤a≤n, 1≤b≤n, a̸= b) which describe that A and b Is friends. There could be repetitions among the given pairs.
Output
The output for each test case should contain (on a line by itself) one number denoting how many people there is in the LA Rgest Group of Friends on a line by itself.
Sample Input
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 1
1 2
9 10
8 9
Sample Output
3
7

English:
A city has n inhabitants, these residents have friends relationship, a friend of a friend can also count as friends, finally asked you this n people, the most friends of the community has how many people.

#include <bits/stdc++.h> using namespace std;
int father[30001];
int rank[30001];
    int Find (int x) {if (x==father[x]) return x;
Return Father[x]=find (Father[x]);
    } void Merge (int x,int y) {x=find (x);
    Y=find (y);
    if (x==y) return;
        if (Rank[x]<=rank[y]) {father[x]=y;
    RANK[Y]+=RANK[X];
        } else {father[y]=x;
    Rank[x]+=rank[y];
        }} void init (int n) {for (int i=1;i<=n;i++) {father[i]=i;
    Rank[i]=1;
    }} int main () {Ios::sync_with_stdio (false);
    int n,m,t;
    cin>>t;
        while (t--) {cin>>n>>m;
        Init (n);
            for (int i=1;i<=m;i++) {int A, B;
            cin>>a>>b;
        Merge (A, b);
        } int ans=-1;
        for (int i=1;i<=n;i++) {Ans=max (ans,rank[i]);
    } cout<<ans<<endl;
} return 0;



 }

Answer:

Naked and check set, in rank among the number of friends to retain the first person, the merger when the number of friends to add to another person can.