HDU——T 2818 Building Block

Posted 2020-10-03

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、、

 1 #include <algorithm>
 2 #include <cstring>
 3 #include <cstdio>
 4 
 5 using namespace std;
 6 
 7 const int N(30005);
 8 int fa[N],sum[N],beh[N];
 9 
10 int find(int x)
11 {
12     if(fa[x]==x) return x;
13     int dad=fa[x];
14     fa[x]=find(fa[x]);
15     beh[x]+=beh[dad];
16     return fa[x];
17 }
18 void combine(int x,int y)
19 {
20     x=find(x),y=find(y);
21     if(x==y) return ;
22     beh[x]=sum[y];
23     sum[y]+=sum[x];
24     fa[x]=y;
25 }
26 void init()
27 {
28     for(int i=0;i<N;i++) fa[i]=i,sum[i]=1,beh[i]=0;
29 }
30 
31 int main()
32 {
33     for(int p,u,v;~scanf("%d",&p);)
34     {
35         init();
36         for(char ch[2];p--;)
37         {
38             scanf("%s%d",ch,&u);
39             if(ch[0]==‘M‘)
40             {
41                 scanf("%d",&v);
42                 combine(u,v);
43             }
44             else find(u),printf("%d\n",beh[u]);
45         }
46     }
47     return 0;
48 }