poj 2524 求连通分量（并查集模板题）

求连通分量

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

 1 # include <iostream>
 2 # include <cstdio>
 3 # include <cstring>
 4 # include <algorithm>
 5 # include <cmath>
 6 # include <queue>
 7 # define LL long long
 8 using namespace std ;
 9
10 const int MAXN = 50010;
11 int F[MAXN];
12 int num[MAXN] ;
13
14 int find(int x)//找x的祖先结点
15 {
16     if(F[x]==x) return x;
17     return F[x]=find(F[x]);
18 }
19 void bing(int u,int v) //按秩合并
20 {
21     int x = find(u);
22     int y = find(v);
23     if(x == y)
24         return ;
25     if(num[x] >= num[y])
26     {
27         F[y] = x;
28         num[x] += num[y];
29     }
30     else
31     {
32         F[x] = y;
33         num[y] += num[x];
34     }
35 }
36 int main()
37 {
38     //freopen("in.txt","r",stdin) ;
39     int n , m ;
40     int Case = 0 ;
41     while(scanf("%d %d", &n , &m) != EOF)
42     {
43          Case++ ;
44          if (n == 0 && m == 0)
45              break ;
46
47          int i ;
48          for(i = 1 ; i <= n ; i++)
49          {
50              F[i] = i ;
51              num[i] = 1 ;
52          }
53         int u , v ;
54         while(m--)
55         {
56             scanf("%d %d" , &u , &v) ;
57             bing(u , v) ;
58
59         }
60         int res = 0 ;
61         for(i = 1 ; i <= n ; i++)
62             if (F[i] == i)
63                res++ ;
64         printf("Case %d: %d\n" , Case , res) ;
65     }
66     return 0;
67 }