poj.1988.Cube Stacking(并查集)

Cube Stacking

Time Limit:2000MS     Memory Limit:30000KB     64bit IO Format:%I64d & %I64u

Submit Status Practice POJ 1988

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

 #include<stdio.h>
 #include<string.h>
 const int M = 3e4 +  ;
 int f[M] , up[M] , tot[M] ;
 int p ;
 int s[] ;
 int u , v ;

 int find (int u)
 {
     if (f[u] == u) return f[u] ;
     int t = f[u] ;
     f[u] = find (f[u]) ;
     up[u] += up[t] ;
     return f[u] ;
 }

 void Union (int u , int v)
 {
     int _u = find (u) , _v = find (v) ;
     if (_u != _v) {
         f[_v] = _u ;
         up[_v] += tot[_u] ;
         tot[_u] += tot[_v] ;
     }
 }

 int main ()
 {
     //freopen ("a.txt" , "r" , stdin ) ;
     for (int i =  ; i < M ; i ++) {
         f[i] = i ;
         up[i] =  ;
         tot[i] =  ;
     }
     scanf ("%d" , &p) ;
     while (p --) {
         scanf ("%s" , s) ;
         if(s[] == 'M') {
             scanf ("%d%d" , &u , &v) ;
             Union (u , v) ;
         }
         else if (s[] == 'C') {
             scanf ("%d" , &u) ;
             int _u = find (u) ;
             printf ("%d\n" , tot[_u] - up[u] - ) ;
         }
     }
     return  ;
 }

一些情况：