Cube Stacking
| Time Limit: 2000MS | Memory Limit: 30000K | |
| Total Submissions: 28148 | Accepted: 9879 | |
| 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
Source File
#include<iostream>
#include<cstdio>
using namespace std;
#define MAXN 30000
int parent[MAXN+5];
int sum[MAXN+5];
int under[MAXN+5];
int getRoot(int a){
int t;
if(parent[a]==a)
return a;
t=getRoot(parent[a]);
under[a]+=under[parent[a]];
parent[a]=t;
return parent[a];
}
void Merge(int a,int b){
int pa=getRoot(a);
int pb=getRoot(b);
if(pa==pb)
return ;
parent[pb]=pa;
under[pb]=sum[pa];
sum[pa]+=sum[pb];
}
int main(){
int p;
cin>>p;
for(int i=0;i<MAXN;i++){
parent[i]=i;
sum[i]=1;
under[i]=0;
}
for(int i=0;i<p;i++){
char s[2];
scanf("%s",s);
int a,b;
if(s[0]=='M'){
scanf("%d%d",&a,&b);
Merge(b,a);
}
else{
scanf("%d",&a);
getRoot(a);
printf("%d\n",under[a]);
}
}
}