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("%I64dn",under[a]);
61       }
62 
63      }
64 
65  }
66  return 0;
67 }

hdu 2818 Building Block

Building Block