PKU POJ 2524 解题报告(并查集)

文章作者：yx_th000 文章来源：Cherish_yimi (http://www.cnblogs.com/cherish_yimi/) 转载请注明，谢谢合作。

并查集学习--并查集详解
 

Ubiquitous Religions

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 9637		Accepted: 4463

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

我的思路：

有了前面学习的基础和1161的练习，这道题完全就可以水过了，设定一个计数器，每次合并时减1便可值得一提的是：树的秩是好东西啊，既可以记录每个集合的节点个数，又能保证按秩合并成相对高度小的树。

代码如下：

   
     

    
      
    
      
    
      Code
    
      
 
        1#include<iostream>
 2using namespace std;
 3
 4int n, m, maxNum, i;
 5int father[50005], num[50005];
 6
 7void makeSet(int n)
 8{
 9    for(int j = 1; j <= n; j++)
10    {
11        father[j] = j;
12        num[j] = 1;
13    }
14}
15int findSet(int x)
16{
17    if(father[x] != x) 
18    {    
19        father[x] = findSet(father[x]);
20    }
21    return father[x]; 
22}
23
24void Union(int a, int b)
25{
26    int x = findSet(a);
27    int y = findSet(b);
28    if(x == y)
29    {
30        return;
31    }
32    if(num[x] <= num[y])
33    {
34        father[x] = y;
35        num[y] += num[x];
36        maxNum--;
37    }
38    else 
39    {
40        father[y] = x;
41        num[x] += num[y];
42        maxNum--;
43    }
44}
45
46int main()
47{
48    int Case = 1;
49    while(scanf("%d %d", &n, &m)!=EOF && n!=0)
50    {
51        maxNum = n;
52        makeSet(n);
53        int a, b;
54        for(i = 0; i < m; i++)
55        {
56            scanf("%d %d",&a, &b);
57            Union(a,b);
58        }
59        printf("Case %d: %d\n",Case++,maxNum);
60    }
61    return 0;
62}
63