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