poj 1988 Cube Stacking

Cube Stacking

Time Limit: 2000MS		Memory Limit: 30000K
Total Submissions: 22163		Accepted: 7777
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

USACO 2004 U S Open

题意：有两种操作M 把X堆得碟子放到Y堆上 C X下面有几个碟子

带权并查集，以x为根结点的所有元素减去以X到根结点的距离再减一

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int i,pre[30100];
int vis[30100];//x到根节点的距离 
int dis[30100];//以X为根节点元素的个数 
void init()
{
	for(i=1;i<=30100;i++)
	{
		pre[i]=i;
		vis[i]=0;
		dis[i]=1;
	}
}
int find(int x)
{
	if(x==pre[x])
	return x;
	else
	{
		int root=find(pre[x]);
		vis[x]=vis[x]+vis[pre[x]];
		return pre[x]=root;
	}
 } 
 void merge(int x,int y)
 {
 	int fx=find(x);
 	int fy=find(y);
 	pre[fy]=fx;
 	vis[fy]=dis[fx];
 	dis[fx]+=dis[fy];
 }
 int main()
 {
 	int t,a,b;
 	char str[10];
 	while(~scanf("%d",&t))
 	{
 		init();
 		while(t--)
 	{
 		scanf("%s",str);
 		if(str[0]=='M')
 		{
 			scanf("%d%d",&a,&b);
 			merge(a,b);
		 }
		 else
		 {
		 	scanf("%d",&a);
		 	int fx=find(a);
		 	printf("%d\n",dis[fx]-vis[a]-1);
		 }
	 }
	 }
 	return 0;
 }