hdu 2818 Building Block

Building Block

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3250    Accepted Submission(s): 973

Problem Description

John are playing with blocks. There are N blocks (1 <= N <= 30000) numbered 1...N。Initially, there are N piles, and each pile contains one block. Then John do some operations P times (1 <= P <= 1000000). There are two kinds of operation:

M X Y : Put the whole pile containing block X up to the pile containing Y. If X and Y are in the same pile, just ignore this command. 
C X : Count the number of blocks under block X 

You are request to find out the output for each C operation.

 

 

Input

The first line contains integer P. Then P lines follow, each of which contain an operation describe above.

 

 

Output

Output the count for each C operations in one line.

 

 

Sample Input

6 M 1 6 C 1 M 2 4 M 2 6 C 3 C 4

 

 

Sample Output

1 0 2

 

 

Source

2009 Multi-University Training Contest 1 - Host by TJU

 

题意：

     1 /*有N个砖头， 编号1—N(实际上是0-N)，然后有两种操作，第一种是M x y 把x所在的那一堆砖头全部移动放到y所在的那堆上面。 第二种操作是 C x ，即查询x下面有多少个砖头 ，并且输出。 2 */

 

 

--->带权值的并查集 

     代码：

 

 

 

 1 #include<cstring>

 2 #include<cstdio>

 3 #include<cstdlib>

 4 #define maxn 30030

 5 using namespace std;

 6 int father[maxn];

 7 __int64 rank[maxn],under[maxn];

 8 int p;

 9 

10  void init(){

11 

12   for(int i=0;i<maxn ;i++) {

13       father[i]=i;

14       rank[i]=1;

15       under[i]=0;

16   }

17 

18 }

19 

20 int fin(int x){

21 

22     if(x  ==  father[x])

23         return  father[x];

24     int tem = father[x] ;

25     father[x] = fin(father[x]);

26      under[x]+=under[tem];

27 

28     return  father[x];

29 

30 }

31 

32 void Union(int a,int b){

33 

34     int x = fin(a);

35     int y = fin(b);

36     if( x==y ) return ;

37     father[x] = y ;       //将a所在的堆放在b的堆上

38     under[x] = rank[y];

39     rank[y] += rank[x];

40     rank[x] = 0;

41 

42 }

43 

44 int main()

45 {

46     char str[2];

47     int a,b;

48 

49  while(scanf("%d",&p)!=EOF){

50     init();

51      while(p--){

52       scanf("%s",str);

53       if(str[0]=='M'){

54           scanf("%d%d",&a,&b);

55           Union(a,b);

56       }

57       else{

58         scanf("%d",&a);

59         fin(a);

60         printf("%I64d\n",under[a]);

61       }

62 

63      }

64 

65  }

66  return 0;

67 }