Cube Stacking（带权并查集）

Cube Stacking

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

这道题的大意是当发出M指令时，进行操作，即将n栈放到m栈上方，当发出C指令时，输出n下方栈的个数

这道题是今天难度比较大的一题，而且不能用cin cout除非关掉同步流，否则会超时。

在find函数更新父结点时，标记n结点的大小，和与最上面结点的距离，输出即为堆栈的个数减去距离最上面结点的距离，再减去自身，即为该栈下面栈的个数。

ac代码：

#include<cstdio>
#include<algorithm>
#include<string.h>
using namespace std;

int pre[30005],cnt[30005],dis[30005];

int find(int n){
	if(pre[n]==n) return n;
	int temp=find(pre[n]);
	dis[n]+=dis[pre[n]];
	pre[n]=temp;
	return temp;
}

void Union(int x,int y){
	int fx=find(x),fy=find(y);
	if(fx==fy) return;
	pre[fy]=fx;
	dis[fy]=cnt[fx];
	cnt[fx]+=cnt[fy];
}

int main(){
	int n,u,v;
	char q[3];
	while(~scanf("%d",&n)){
		memset(dis,0,sizeof(dis));
		for(int i=1;i<=30005;i++){
			cnt[i]=1;
			pre[i]=i;
		}
		for(int i=0;i<n;i++){
			scanf("%s",q);
			if(q[0]=='M'){
				scanf("%d %d",&u,&v);
				Union(u,v);
			}
			else{
				scanf("%d",&u);
				int x=find(u);
				//cout<<x<<" "<<cnt[x]<<" "<<dis[u]<<endl;
				printf("%d\n",cnt[x]-dis[u]-1);
			}
		}
	}
	return 0;
}