cf Educational Codeforces Round 65 C. News Distribution

原題：

C. News Distribution
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

In some social network, there are n users communicating with each other in m groups of friends. Let’s analyze the process of distributing some news between users.

Initially, some user x receives the news from some source. Then he or she sends the news to his or her friends (two users are friends if there is at least one group such that both of them belong to this group). Friends continue sending the news to their friends, and so on. The process ends when there is no pair of friends such that one of them knows the news, and another one doesn’t know.

For each user x you have to determine what is the number of users that will know the news if initially only user x starts distributing it.

Input

The first line contains two integers n and m (1≤n,m≤5⋅10^5) — the number of users and the number of groups of friends, respectively.

Then m lines follow, each describing a group of friends. The i-th line begins with integer ki (0≤ki≤n) — the number of users in the i-th group. Then ki distinct integers follow, denoting the users belonging to the i-th group.

It is guaranteed that ∑i=1mki≤5⋅10^5

.
Output

Print n
integers. The i-th integer should be equal to the number of users that will know the news if user i

starts distributing it.
Example
Input

7 5
3 2 5 4
0
2 1 2
1 1
2 6 7

Output

4 4 1 4 4 2 2

中文：

有n個人，每個人屬於一些社交圈，一個社交圈的裏的人會傳播消息。現在給處社交圈，問你每個人作爲消息傳播源的情況下，對應知道消息的人數。

代碼:

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
 
typedef pair<int,int> pii;
const int maxn = 500001;
int father[maxn];
int Rank[maxn];
int n,m;
void init()
{
    for(int i=1;i<=n;i++)
    {
        father[i]=i;
        Rank[i]=1;
    }
}
 
 
int Find(int x)
{
    if(x!=father[x])
        return father[x] = Find(father[x]);
    else
        return x;
}
 
void unite(int x,int y)
{
    x = Find(x);
    y = Find(y);
    if(x==y)
        return;
    if(Rank[x]<Rank[y])
    {
        swap(x,y);
    }
    father[y]=x;
    Rank[x]+=Rank[y];
}
 
 
int main()
{
    ios::sync_with_stdio(false);
    while(cin>>n>>m)
    {
        init();
        int c;
        for(int i=1;i<=m;i++)
        {
            cin>>c;
            int mark;
            if(c)
            {
                int val;
                cin>>mark;
                for(int j=1;j<c;j++)
                {
                    cin>>val;
                    unite(mark,val);
                }
            }
        }
        for(int i=1;i<=n;i++)
        {
            int x = Find(i);
            if(i!=n)
            cout<<Rank[x]<<" ";
            else
                cout<<Rank[x]<<endl;
        }
 
    }
    return 0;
}

解答：
裸的並查集，把按秩合併的部分改成人數就完事了