# Box

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

## Problem Description

There are N boxes on the ground, which are labeled by numbers from 1 to N. The boxes are magical, the size of each one can be enlarged or reduced arbitrarily. Jack can perform the “MOVE x y” operation to the boxes: take out box x; if y = 0, put it on the ground; Otherwise, put it inside box y. All the boxes inside box x remain the same. It is possible that an operation is illegal, that is, if box y is contained (directly or indirectly) by box x, or if y is equal to x. In the following picture, box 2 and 4 are directly inside box 6, box 3 is directly inside box 4, box 5 is directly inside box 1, box 1 and 6 are on the ground.

The picture below shows the state after Jack performs “MOVE 4 1”:

Then he performs “MOVE 3 0”, the state becomes:

During a sequence of MOVE operations, Jack wants to know the root box of a specified box. The root box of box x is defined as the most outside box which contains box x. In the last picture, the root box of box 5 is box 1, and box 3’s root box is itself.

## Input

Input contains several test cases. For each test case, the first line has an integer N (1 <= N <= 50000), representing the number of boxes. Next line has N integers: a1, a2, a3, … , aN (0 <= ai <= N), describing the initial state of the boxes. If ai is 0, box i is on the ground, it is not contained by any box; Otherwise, box i is directly inside box ai. It is guaranteed that the input state is always correct (No loop exists). Next line has an integer M (1 <= M <= 100000), representing the number of MOVE operations and queries. On the next M lines, each line contains a MOVE operation or a query:

MOVE x y, 1 <= x <= N, 0 <= y <= N, which is described above. If an operation is illegal, just ignore it.

QUERY x, 1 <= x <= N, output the root box of box x.

## Output

For each query, output the result on a single line. Use a blank line to separate each test case.

## Sample Input

2

0 1

5

QUERY 1

QUERY 2

MOVE 2 0

MOVE 1 2

QUERY 1

6

0 6 4 6 1 0

4

MOVE 4 1

QUERY 3

MOVE 1 4

QUERY 1

## Sample Output

1

1

2

1

1

## 题意

动态地维护一些盒子套盒子的操作，询问根。

## 分析

盒子与盒子的关系可以直观地用树的结构来表示，一个结点下的子结点可以表示大盒子里面直接套着的小盒子。

所以本题就是一个裸的Link-Cut Tree模型了。

关于LCT树，还是推荐Yang Zhe的QTREE论文吧。

动态树是用访问操作来划分树链，对于每一条树链，使用Splay来维护，用深度作为splay的左右关系。

看了很多代码，觉得还是写不好，总觉得别人的用起来不顺，最后是在自己原来Splay的基础上改的。

原本的整棵树是个splay，但是在LCT中，整棵树是由很多棵分散的Splay组合起来的，于是在其中的一些点上加上root标记，表示以这一点为根下面可以形成一棵splay树。多个这样的splay组合完成之后就是一棵LCT了。

后面的代码中加入了输入输出挂。。。。。。

1 | /* *********************************************** |