HDU——T 2818 Building Block

http://acm.hdu.edu.cn/showproblem.php?pid=2818

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5935    Accepted Submission(s): 1838

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

 

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多组数据、、貌似因为这个WA好多次了、、

 #include <algorithm>
 #include <cstring>
 #include <cstdio>

 using namespace std;

 const int N();
 int fa[N],sum[N],beh[N];

 int find(int x)
 {
     if(fa[x]==x) return x;
     int dad=fa[x];
     fa[x]=find(fa[x]);
     beh[x]+=beh[dad];
     return fa[x];
 }
 void combine(int x,int y)
 {
     x=find(x),y=find(y);
     if(x==y) return ;
     beh[x]=sum[y];
     sum[y]+=sum[x];
     fa[x]=y;
 }
 void init()
 {
     for(int i=;i<N;i++) fa[i]=i,sum[i]=,beh[i]=;
 }

 int main()
 {
     for(int p,u,v;~scanf("%d",&p);)
     {
         init();
         for(char ch[];p--;)
         {
             scanf("%s%d",ch,&u);
             if(ch[]=='M')
             {
                 scanf("%d",&v);
                 combine(u,v);
             }
             else find(u),printf("%d\n",beh[u]);
         }
     }
     return ;
 }