# 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

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 }```

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