并查集--The Suspects

The Suspects

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

首先先定义数组：fa[]表示这一个点的父亲节点，

并查集一共包含三个部分：

1：初始化     先把所有的点的父亲节点都定义为他本身

1 void init(int n)
2     {
3         for (int i = 0 ;i<=n ;i++)
4         fa[i]=i;
5     }

2： 找根节点   如果此点的父亲节点是他本身，他就是父亲节点，如果不是就找他父亲节点的父亲节点

　　　　　　　直到他的父亲节点是他本身

1 int find(int v)
2 {
3     if (v==fa[v])
4     return v;
5     fa[v]=find(fa[v]);
6     return fa[v];
7     }

3. 合并  把一个数的父亲节点指向另一个节点即把他们两个合并

1 void update(int u,int v)
2 {
3     int fu= find(u);
4     int fv= find(v);
5     fa[fu] =fv; 
6 } 

把每一组数据中的人，都合并起来（参照并查集模版），最后只需要判断从1到n这所有人中是否与0处于1个集合。即fa[i]=fa[0]

 1 #include 
 2 #include 
 3 #include 
 4 #include 
 5 using namespace std;
 6 int n,m;
 7 int k,a[30000];
 8 int fa[30000];
 9 void init(int n)
10     {
11         for (int i = 0 ;i<=n ;i++)
12         fa[i]=i;
13     }
14 int find(int v)
15 {
16     if (v==fa[v])
17     return v;
18     fa[v]=find(fa[v]);
19     return fa[v];
20     }
21 void update(int u,int v)
22 {
23     int fu= find(u);
24     int fv= find(v);
25     fa[fu] =fv; 
26 } 
27 int main()
28 {
29     while (scanf ("%d%d",&n,&m)&&n+m!=0)
30     {
31         int ans=0;
32         memset(a,-1,sizeof(a));
33         init(n);
34         for (int i = 1;i <= m;i++)
35         {
36             scanf ("%d",&k);
37             for (int j = 1;j <= k;j++)
38             {
39                 scanf ("%d",&a[j]);
40                 if (j!=1)
41                 update(a[j],a[j-1]);
42             }
43         }
44         for (int j = 0;j <= n;j++)
45         {
46             if (find(j)==find(0))
47             ans++;
48         }
49         cout<endl;
50     }
51     return 0;
52 }