Wearing masks

Meaning of the questions:

Novel coronavirus pneumonia (Corona Virus Disease 2019, COVID- 19), referred to as "the new crown pneumonia" refers to the 2019 novel coronavirus infection caused by pneumonia.
If an infected person into a group, then this group need to be isolated!
A small student was diagnosed with the new crown infection, and does not wear a mask! ! ! ! ! !
Risk! ! !
Time is running out! ! ! !
Need to find as quickly as possible to all students and small A direct or indirect contact through the students, they will be isolated to prevent the spread of a wider range.
As we all know, students' communication may be in small groups, a student may also participate in multiple small groups.
Please write a program to solve! wear mask! !

Input

A plurality of sets of data, for each set of test data:
the first two integers n and Behavior m (n = m = 0 indicates the end of input, does not need to process), n is the number of students, m is the number of groups of students. 0 <n <= 3e4, 0 <= m <= 5e2
student number of 0 ~ n-1
A small number 0.
Then, m lines, each person has a number of small groups i.e. integer num. Followed by a num integer representing students in small groups.

Output

The number of output to be isolated, the answer to the output of each data per line

Sample Input

100 4
2 1 2
5 10 13 11 12 14
2 0 1
2 99 2
200 2
1 5
5 1 2 3 4 5
1 0
0 0

Sample Output

4
1
1

My Solution:

 Disjoint-set : to "category" element, to a certain element as a representative element, form a set of different categories.

Charles: Find the corresponding representatives yuan to determine a property

And: by checking whether the same kind of element is determined, and then determines whether or not combined into one earth element.

Set: The several elements are bundled together to form a set of representatives of dollars.

This question need only count the number of groups working with the source of infection associated with a simple disjoint-set problems.

 1 #include<iostream>
 2 #include<stdio.h>  
 3 #include<algorithm>
 4 #include<string.h>
 5 using namespace std;
 6 
 7 const int maxn=100005; 
 8 int n,maxL,maxlen,f1,f2;
 9 int L1[maxn],L2[maxn];
10 int u,w;
11 typedef struct node{
12     int u,v,w,nxt;
13 }Edge;
14 Edge Edges[maxn];
15 int head[maxn],tot,vis[maxn];
16 
17 void addEdge(int u,int v,int w){
18     Edges[tot].u=u;
19     Edges[tot].v=v;
20     Edges[tot].w=w;
21     Edges[tot].nxt=head[u];
22     head[u]=tot;
23     tot++;
24 }
25 
26 void dfs(int u,int *a,int L){  
27    a[u]=L; 
28    if(maxL<L) {
29        maxL=L;maxlen=u;
30    }
31    for(int i=head[u];i!=-1;i=Edges[i].nxt){
32       if(!vis[Edges[i].v]){ 
33           vis[Edges[i].v]=true; 
34           dfs(Edges[i].v, a, L + Edges[i].w);
35       }
36   } 
37 } 
38 int main(){
39     while(cin>>n){
40         
41     tot=0;
42     for(int i=0;i<n+2;++i) head[i]=-1;
43     
44     for(int i=2;i<=n;++i){
45         cin>>u>>w;
46         addEdge(i,u,w);
47         addEdge(u,i,w); //加入双向边 
48     } 
49     
50     memset(vis,0,sizeof(vis)); 
51     maxL=-1;vis[1]=1;
52     
53     dfs(1 , L 1, 0 ); 
54      F1 = maxlen;
55      memset (view, 0 , sizeof (show));
56      maxL = - 1 ;
57      show [f1] = 1 ;
58      
59      DFS (F1, L1, 0 ); 
60      f2 = maxlen;
61      memset (view, 0 , sizeof (show));
62      show [f 2] = 1 ;
63      DFS (F2, L2, 0 );
64      
65      for ( int= I . 1 ; I <= n-; I ++ ) {
 66          COUT << max (Ll [I], L2 of [I]) << endl;
 67          // greater is the output 
68        }    
 69    }
 70      return  0 ;
 71 is }