Codeforces 1167c(ccpc wannafly camp day1) News Distribution 并查集模板

题目：

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

Initially, some user xx 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 xx you have to determine what is the number of users that will know the news if initially only user xx starts distributing it.

输入：

The first line contains two integers nn and mm (1≤n,m≤5⋅1051≤n,m≤5⋅105) — the number of users and the number of groups of friends, respectively.

Then mm lines follow, each describing a group of friends. The ii-th line begins with integer kiki (0≤ki≤n0≤ki≤n) — the number of users in the ii-th group. Then kiki distinctintegers follow, denoting the users belonging to the ii-th group.

It is guaranteed that ∑i=1mki≤5⋅105∑i=1mki≤5⋅105.

输出：

Print nn integers. The ii-th integer should be equal to the number of users that will know the news if user ii starts distributing it.

样例：

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 

分析：不想分析。。。

 1 #include<iostream>
 2 #include<sstream>
 3 #include<cstdio>
 4 #include<cstdlib>
 5 #include<string>
 6 #include<cstring>
 7 #include<algorithm>
 8 #include<functional>
 9 #include<iomanip>
10 #include<numeric>
11 #include<cmath>
12 #include<queue>
13 #include<vector>
14 #include<set>
15 #include<cctype>
16 const double PI = acos(-1.0);
17 const int INF = 0x3f3f3f3f;
18 const int NINF = -INF - 1;
19 typedef long long ll;
20 #define MOD 1000007
21 using namespace std;
22 int far[500005];
23 int sum[500005];
24 int n, m;
25 int find(int x)
26 {
27     if(far[x] == x) return x;
28     else return far[x] = find(far[x]);
29 }
30 bool check(int x, int y)
31 {
32     return find(x) == find(y);
33 }
34 void unite(int x, int y)
35 {
36     x = find(x), y = find(y);
37     if(x == y) return;
38     far[y] = x;
39     sum[x] += sum[y];
40 }
41 void init(int n)
42 {
43     for(int i = 0;i <= n;i++)
44     {
45         far[i] = i;
46         sum[i] = 1;
47     }
48 }
49 int main()
50 {
51     cin >> n >> m;
52     init(n);
53     int t;
54     int a, b;
55     while (m--)
56     {
57         cin >> t;
58         if (!t) continue;
59         cin >> a;
60         for (int i = 1; i < t; ++i)
61         {
62             cin >> b;
63             if (!check(a, b) )
64             {
65                 unite(a, b);
66             }
67         }
68     }
69     for(int i = 1 ;i <= n; ++i)
70     {
71         int x = find(i);
72         cout << sum[x] << ' ';
73     }
74     return 0;
75 }