L2-024 tribe (25 points) (and check the collection)

describe ：

In a community , Everyone has their own small circle , It may also belong to many different circles of friends at the same time . We think that friends of friends are all included in a tribe , So I want you to count , In a given community , How many different tribes are there ？ And check if any two people belong to the same tribe .

Input format ：

Output ：

First, output the total number of people in this community in one line 、 And the number of tribes that don't intersect each other . Then for each query , If they belong to the same tribe , Output in one line Y, Otherwise output N.
sample input ：

4
3 10 1 2
2 3 4
4 1 5 7 8
3 9 6 4
2
10 5
3 7

sample output ：

10 2
Y
N

 #include<bits/stdc++.h>
using namespace std;
 
typedef unsigned long long ull;
typedef long long ll;
 
const ll maxx = 1e18;
const int N = 1e6+100;
const int p = 1e4+10;
const double eps = 1e-8;

int n,m,t,k;
int a,bs;
int b[p];
int pre[N],max1,cnt;
map<int,int>mp;

int find(int x)
{
    
	if(x==pre[x]) return x;
	else return pre[x]=find(pre[x]);
}

void unionn(int x,int y)
{
    
	int ya=find(x);
	int yb=find(y);
	pre[ya]=yb;
}

int main()
{
    
	
	cin>>n;
	
	for(int i=1;i<=1e6;i++)
	{
    
		pre[i]=i;
	}// Processing the root node 
	
	for(int i=1;i<=n;i++)
	{
    
		cin>>k;
		for(int j=1;j<=k;j++)
		{
    
			cin>>b[j];
			max1=max(max1,b[j]);
		}// Input , Find out the total number of people 
		
		for(int j=1;j<=k;j++)
		{
    
			for(int s=j+1;s<=k;s++)
			{
    
				unionn(b[j],b[s]);
			}
		}// and 
	}
	
	cin>>m;
	
	for(int i=1;i<=max1;i++)
	{
    
		if(mp[find(i)]==0)
		{
    
			mp[find(i)]=1;
			cnt++;
		}// Number of intersecting sets not found 
	}
	
	cout<<max1<<" "<<cnt<<endl;
	
	for(int i=1;i<=m;i++)
	{
    
		cin>>a>>bs;
		
		if(find(a)==find(bs))
		cout<<"Y"<<endl;
		else
		cout<<"N"<<endl;
	}// check 
	

	
	return 0;
}