个人心得：最基础的并查集经典题。借此去了解了一下加深版的即加权并查集，比如食物链的题目，这种题目实行起来还是有
一定的难度，不仅要找出与父节点的关系，还要在路径压缩的时候进行更新，这一点现在还是没那么上手，不过先知道思维
还是好的吧。这道水题就不多提了...就是注意合并的时候以数小的为跟节点就好了
Severe acute respiratory syndrome (SARS), an atypical pneumonia of unknown aetiology, was recognized as a global threat in mid-March 2003. To minimize transmission to others, the best strategy is to separate the suspects from others.
In the Not-Spreading-Your-Sickness University (NSYSU), there are many student groups. Students in the same group intercommunicate with each other frequently, and a student may join several groups. To prevent the possible transmissions of SARS, the NSYSU collects the member lists of all student groups, and makes the following rule in their standard operation procedure (SOP).
Once a member in a group is a suspect, all members in the group are suspects.
However, they find that it is not easy to identify all the suspects when a student is recognized as a suspect. Your job is to write a program which finds all the suspects.
Input
The input file contains several cases. Each test case begins with two integers n and m in a line, where n is the number of students, and m is the number of groups. You may assume that 0 < n <= 30000 and 0 <= m <= 500. Every student is numbered by a unique integer between 0 and n?1, and initially student 0 is recognized as a suspect in all the cases. This line is followed by m member lists of the groups, one line per group. Each line begins with an integer k by itself representing the number of members in the group. Following the number of members, there are k integers representing the students in this group. All the integers in a line are separated by at least one space.
A case with n = 0 and m = 0 indicates the end of the input, and need not be processed.
Output
For each case, output the number of suspects in one 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
1 #include<iostream>
2 #include<cstdio>
3 #include<cstring>
4 using namespace std;
5 int n,m;
6 int person[30005];
7 void init()
8 {
9 for(int i=0;i<n;i++)
10 person[i]=i;
11 }
12 int geta(int x)
13 {
14 if(person[x]!=x)
15 person[x]=geta(person[x]);
16 return person[x];
17 }
18 void mergea(int x,int y)
19 {
20 if(geta(x)<geta(y))
21 person[geta(y)]=geta(x);
22 else
23 person[geta(x)]=geta(y);
24 }
25 int number(){
26 int sum=0;
27 for(int i=0;i<n;i++)
28 if(geta(i)==0) sum++;
29 return sum;
30 }
31 int main()
32 {
33
34 while(cin>>n>>m)
35 {
36 init();
37 if(!m&&!n) break;
38 while(m--)
39 {
40 int text,x,y;
41 int flag=0;
42 cin>>text;
43 while(text--)
44 {
45 cin>>x;
46 if(flag==0)
47 {flag=1;y=x;}
48 else
49 mergea(y,x);}
50
51 }
52 int sum=number();
53 cout<<sum<<endl;
54
55
56 }
57 return 0;
58 }