Cube Stacking

#include <stdio.h>
int father[30011],deep[30011],h[30011];//数组H[i]表示i下有多少个方块 ，deep[i]表示i所在栈堆的高度 
int find(int x)
{
	if(x!=father[x])
	{
		int t=father[x];
		father[x]=find(father[x]);
		h[x]+=h[t]; 
	}
	return father[x];
}
void join(int x, int y)
{
	  x=find(x);
	  y=find(y);
	if(x!=y)
	{
		father[x]=y;
		h[x]=deep[y];
		deep[y]+=deep[x];
		deep[x]=0;
		
	}
}
int main()
{
	int n,a,b,p;
	char op[2];
	scanf("%d",&n);
	for(int i=1;i<30001;i++)
	{
		father[i]=i;
		deep[i]=1;
		h[i]=0;
	}
	for(int i=1;i<=n;i++)
	{
		scanf("%s",op);
		if(op[0]=='M')
		{
			scanf("%d%d",&a,&b);
			join(a,b);
		}
		else {
			scanf("%d",&p);
			int k=find(p);
			printf("%d\n",h[p]);
		}
	}
	return 0;
 }

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

Sample Output

1
0
2

此处属于复制：

给定N个方块，排成一行，将它们编号1到N。再给出P个操作：

①M i j表示将i所在的那一堆移到j所在那一堆的顶上。

②C i表示一个询问，询问i下面有多少个方块。

•你需要写一个程序来完成这些操作。

我们设f[i]是每个方块所在位置的最底下方块的编号，h[i]是每个方块之下有多少个方块（开始赋成0）。每次读入合并操作Y和X时：

首先找到X和Y的父亲（最底下的点），同时更新：f[fy]=fx。但是很快问题就来了——我们怎么更新f[fy]的高度呢？我们不能获知左边堆里最上面一层的h[i]!

咨询了RZZ后，我又开了个数组deep[i],表示i所在的堆内的高度（仅当i是底层的方格）。初始化时是1。进行如图的操作时,h[fy]=deep[fx]，同时deep[fx]+=deep[fy]。

有人可能要问：右侧堆中底层的元素已经转移了，那么它上面的点呢？这里，要用到类似于线段树里的lazy-tag思想，即询问时再更新值。如果询问右侧某一点，我们只需在它寻找父亲、路径压缩时，再增加一个修改h的操作即可。

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