POJ——T 1988 Cube Stacking

http://poj.org/problem?id=1988

Time Limit: 2000MS		Memory Limit: 30000K
Total Submissions: 25865		Accepted: 9044
Case Time Limit: 1000MS

Description

Farmer John and Betsy are playing a game with N (1 <= N <= 30,000)identical cubes labeled 1 through N. They start with N stacks, each containing a single cube. Farmer John asks Betsy to perform P (1<= P <= 100,000) operation. There are two types of operations: 
moves and counts. 
* In a move operation, Farmer John asks Bessie to move the stack containing cube X on top of the stack containing cube Y. 
* In a count operation, Farmer John asks Bessie to count the number of cubes on the stack with cube X that are under the cube X and report that value.

Write a program that can verify the results of the game.

Input

* Line 1: A single integer, P

* Lines 2..P+1: Each of these lines describes a legal operation. Line 2 describes the first operation, etc. Each line begins with a 'M' for a move operation or a 'C' for a count operation. For move operations, the line also contains two integers: X and Y.For count operations, the line also contains a single integer: X.

Note that the value for N does not appear in the input file. No move operation will request a move a stack onto itself.

Output

Print the output from each of the count operations in the same order as the input file. 

Sample Input

6
M 1 6
C 1
M 2 4
M 2 6
C 3
C 4

Sample Output

1
0
2

Source

USACO 2004 U S Open

 

查询x时，先维护x下边的、

 #include <algorithm>
 #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=find(fa[x]);
     beh[x]+=beh[fa[x]];
     return fa[x]=dad;
 }
 inline 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;
 }

 int main()
 {
     for(int i=;i<N;i++)
         fa[i]=i,sum[i]=;
     int p,u,v; scanf("%d",&p);
     for(char ch;p--;)
     {
         scanf("\n%c%d",&ch,&u);
         if(ch=='M')
         {
             scanf("%d",&v);
             combine(u,v);
         }
         else find(u),printf("%d\n",beh[u]);
     }
     return ;
 }