Cube Stacking
| Time Limit: 2000MS | Memory Limit: 30000K | |
| Total Submissions: 21157 | Accepted: 7395 | |
| 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.
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>
#define N 30001 int count[N], num[N], pre[N]; void inite()
{
for(int i = ; i < N; i++)
{
count[i] = ;
num[i] = ;
pre[i] = i;
}
} int find(int x)
{
if(pre[x] == x)
return x; int t = find(pre[x]);
count[x] += count[pre[x]];
pre[x] = t;
return t; }
void Union(int x, int y)
{
int i = find(x);
int j = find(y);
if(i == j)
{
return;
}
count[i] = num[j];
num[j] += num[i];
pre[i] = j;
} int main()
{
int i, x, y, n;
char s[];
scanf("%d",&n);
inite();
for(i = ; i < n; i++)
{
scanf("%s",s);
if(s[] == 'M')
{
scanf("%d%d",&x,&y);
Union(x,y);
}
else if(s[] == 'C')
{
scanf("%d",&x);
int c = find(x);
printf("%d\n",count[x]);
}
}
return ;
}