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：

/*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。 */

--->带权值的并查集

代码：

 #include<cstring>
 #include<cstdio>
 #include<cstdlib>
 #define maxn 30030
 using namespace std;
 int father[maxn];
 __int64 rank[maxn],under[maxn];
 int p;

  void init(){

   for(int i=;i<maxn ;i++) {
       father[i]=i;
       rank[i]=;
       under[i]=;
   }

 }

 int fin(int x){

     if(x  ==  father[x])
         return  father[x];
     int tem = father[x] ;
     father[x] = fin(father[x]);
      under[x]+=under[tem];

     return  father[x];

 }

 void Union(int a,int b){

     int x = fin(a);
     int y = fin(b);
     if( x==y ) return ;
     father[x] = y ;       //将a所在的堆放在b的堆上
     under[x] = rank[y];
     rank[y] += rank[x];
     rank[x] = ;

 }

 int main()
 {
     char str[];
     int a,b;

  while(scanf("%d",&p)!=EOF){
     init();
      while(p--){
       scanf("%s",str);
       if(str[]=='M'){
           scanf("%d%d",&a,&b);
           Union(a,b);
       }
       else{
         scanf("%d",&a);
         fin(a);
         printf("%I64d\n",under[a]);
       }

      }

  }
  return ;
 }