POJ 1988 Cube Stocing

タイトルのソース:http://poj.org/problem?id=1988
Cube Stocing
Time Limit: 2000 MS
 
メモリLimit: 30000 K
Total Submissions: 19173
 
Acceepted: 6693
Case Time Limit: 1000 MS
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 Bets Bets Betsy to perform（1=Theropere of） 
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 result of the game. 
Input
*Line 1:A single integer、P 
Line 2 describes the first operation.Line 2 describes the first operation，etc.Each line begins with a'M'fora move operatior a'C'fora count count operation.For move operininininininininininininininininininininininininininininininininininininininininininininininininininininininininininininininininininininininininininininininininininininininininininininininininininininininininininininininininininininger:X. 
Note that the value for N does not appar in the input file.No move operation will request a move a stack onto itself. 
Output
Print the output from each of the count operation s 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

ソurce
USACO 2004 U S Open
ベクトルを調べて~ Vijos 1443銀河英雄伝説と同じです。今回は本当にさわやかで、一回打っても何の修正もなくACになりました。
ACコード:

#include<cstdio>
const int Max=30001;
int father[Max],loc[Max],root[Max],t,x,y;
char ch;
int find(int k){
	if(k==father[k]) return k;
	int temp=find(father[k]);
	loc[k]+=loc[father[k]];
	father[k]=temp;
	return temp;
}
int main(){
	scanf("%d",&t);
	for(int i=1;i<Max;i++){
		father[i]=i;loc[i]=0;root[i]=1;
	}
	while(t--){
		getchar();
		scanf("%c",&ch);
	if(ch=='M'){
		scanf("%d%d",&x,&y);
		int tempx=find(x),tempy=find(y);
		father[tempy]=tempx;  
		loc[tempy]=root[tempx];
		root[tempx]+=root[tempy];
	}
	else {
		scanf("%d",&x);
		int tempx=find(x);
		printf("%d
",root[tempx]-loc[x]-1);
	}
	}
	return 0;
}