POJ 2524 -- Ubiquitous Religions

http://poj.org/problem?id=2524

Description

There are so many different religions in the world today that it is difficult to keep track of them all. You are interested in finding out how many different religions students in your university believe in.

You know that there are n students in your university (0 < n <= 50000). It is infeasible for you to ask every student their religious beliefs. Furthermore, many students are not comfortable expressing their beliefs. One way to avoid these problems is to ask m (0 <= m <= n(n-1)/2) pairs of students and ask them whether they believe in the same religion (e.g. they may know if they both attend the same church). From this data, you may not know what each person believes in, but you can get an idea of the upper bound of how many different religions can be possibly represented on campus. You may assume that each student subscribes to at most one religion.

Input

The input consists of a number of cases. Each case starts with a line specifying the integers n and m. The next m lines each consists of two integers i and j, specifying that students i and j believe in the same religion. The students are numbered 1 to n. The end of input is specified by a line in which n = m = 0.

Output

For each test case, print on a single line the case number (starting with 1) followed by the maximum number of different religions that the students in the university believe in.

Sample Input

10 9
1 2
1 3
1 4
1 5
1 6
1 7
1 8
1 9
1 10
10 4
2 3
4 5
4 8
5 8
0 0

Sample Output

Case 1: 1
Case 2: 7

http://www.tuicool.com/articles/B7ZzUrv

题目大意：一个学校有n人，其中有m对人是具有相同的宗教信仰，
问这个学校里信仰的宗教数最多有多少个。
解题思路：基础的并查集。具有相同信仰的人归为一个集合，
则最后的集合数量即为结果。求集合数量：判断有多少个根节点，用到根节点的特征(其父节点为自己）

int n,m;
int f[52014];
void init()
{
  for(int i=1; i<=n; i++)
    f[i]=i;
}
int find(int x)
{
  return f[x]==x?x:find(f[x]);
}
void unity(int a,int b)
{
  if(find(a)!=find(b))
    f[find(a)]=find(b);
}
int main()
{
  int ji=1;
  while(cin>>n>>m)
  {
    if(n+m==0)break;
    init();
    while(m--)
    {
      int a,b;
      cin>>a>>b;
      unity(a,b);
    }
    int ans=0;
    for(int i=1; i<=n; i++)
      if(f[i]==i)
        ans++;
    cout<<"Case "<<ji++<<": ";
    cout<<ans<<endl;
  }
  return 0;
}

http://blog.csdn.net/synapse7/article/details/9466269

1. const int maxn = 50001;  
2. int father[maxn], root[maxn];  
3. void Make_Set(int x) {  
4.     father[x] = x;  
5.     root[x] = 1;  
6. }  
7. //路径压缩  
8. int Find_Set(int x) {  
9.     if (father[x] != x) {  
10.         father[x] = Find_Set(father[x]);  
11.     }  
12.     return father[x];  
13. }  
14. void Union_Set(int a, int b) {  
15.     if (root[a] <= root[b]) {  
16.         father[a] = b;  
17.         root[b] += root[a];  
18.     } else {  
19.         father[b] = a;  
20.         root[a] += root[b];  
21.     }  
22. }  
23. int main() {  
24.     int n, m, i, a, b, fa, fb, casenum = 1;  
25.     while (cin >> n >> m && !(n == 0 && m == 0)) {  
26.         for (i = 1; i <= n; i++)  
27.             Make_Set(i);  
28.         for (i = 1; i <= m; i++) {  
29.             cin >> a >> b;  
30.             fa = Find_Set(a);  
31.             fb = Find_Set(b);  
32.             if (fa != fb) {  
33.                 Union_Set(fa, fb);  
34.                 n--;  
35.             }  
36.         }  
37.         printf("Case %d: %d\n", casenum++, n);  
38.     }  
39.     return 0;  
40. }  

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