并查集的应用——Ubiquitous Religions

时间限制: 

5000ms 

内存限制: 

65536kB

描述

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.

输入

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.

输出

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.

样例输入

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

样例输出

Case 1: 1
Case 2: 7

提示

Huge input, scanf is recommended.

1. #include
2. #include
3. using namespace std;
4. int a[30010], StuNum[30010], count;
5. int total, group, eachgroup, i, cal[50010];
6. int min(int x,int y)
7. {
8. if(x>y)
9. return y;
10. else
11. return x;
12. }
13. int getfather(int x)
14. {
15. if(x==a[x])
16. return x;
17. else
18. return getfather(a[x]); ///寻找x所在集合的根结点
19. }
20. int getcount()
21. {
22. int count = 0;
23. for(i=0; i<total; i++)
24. {
25. cal[i]=getfather(i+1);
26. }
27. sort(cal, cal+total);
28. for(i=0; i<total-1; i++)
29. {
30. if(cal[i]!=cal[i+1])
31. count++;
32. }
33. return count;
34. }
35. void combine(int x,int y)
36. {
37. int fx = getfather(x); ///寻找x的根结点
38. int fy = getfather(y); ///寻找y的根结点
39. a[fx] = a[fy] = min(a[fx], a[fy]);
40. }
41. 
    
42. while(cin>>total>>group)
43. int main()
44. {
45. int first, second, asdf=0;
46. {
47. asdf++;
48. if(total==0 && group==0)
49. return 0;
50. else if(group==0)
51. cout<<"Case "<<asdf<<": "<<total<<endl;
52. else
53. {
54. for(i=1; i<=total; i++)
55. a[i] = i;
56. while(group--)
57. {
58. cin>>first>>second;
59. ///此处插入合并代码
60. combine(first, second);///合并完成
61. }
62. ///此处插入搜索代码
63. cout<<"Case "<<asdf<<": "<<getcount()+1<<endl;
64. }
65. }
66. return 0;
67. }