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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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 }