POJ 1988 Cube Stacking

Cube Stacking

Time Limit: 2000MS		Memory Limit: 30000K
Total Submissions: 24452		Accepted: 8563
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

#include<iostream>
#include<cstdio>
#include<cstring>
#define maxn 30001
using namespace std;
int father[maxn],n,m,sum[maxn],cnt[maxn];
int find(int x){
	if(x==father[x]) return x;
	int fa=father[x];
	father[x]=find(father[x]);
	cnt[x]+=cnt[fa];
	return father[x];
}
void megre(int x,int y){// x,y是其本身那一坨的祖宗 
	father[y]=x;
	cnt[y]=sum[x];
	sum[x]+=sum[y];
	sum[y]=0;
}
int main()
{
	for(int i=1;i<=maxn;i++) father[i]=i,sum[i]=1;
	char ch[8];
	int x,y;
	scanf("%d",&m);
	while(m--){
		cin>>ch+1;
		if(ch[1]=='M'){
			scanf("%d%d",&x,&y);
			int rx=find(x),ry=find(y);
			megre(rx,ry);
		}
		else{
			scanf("%d",&x);
			int rx=find(x);
			printf("%d
",sum[rx]-cnt[x]-1);
		}
	}
	return 0;
}

思路：加权并查集例题