2014多校7（1008）hdu4942（树的中序遍历+线段树）

Game on S♂play

Time Limit: 16000/8000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 215    Accepted Submission(s): 36

Problem Description

Uncle Fang is learning Splay. When he heard about the rotate operation, he would like to play a game with Captain Chen. There is a tree with n nodes initially, and the node 1 is the root. The i-th nodes gets a weight of wi, and an attribute of interesting value defined as the sum of the weight of it’s descendants. Following are two operations in this tree:

1.do a rotation on a node x. The rotation is the common rotation in splay tree, it can be right-rotation or left-rotation, as demonstrated in the following picture.If it's impossible to rotate, just ignore it!


2.ask Captain Chen the product of interesting values of the nodes in a subtree rooted at node x.

 

Input

There are several cases.The first line of the input contains a single integer T (T <= 30) which is the number of test cases.Then comes the T test cases .

For each case, the first line contains two integer numbers n and m(1<=n, m<=100000).

The following n lines describe the tree. In the i-th line, there was three number wi, xi, yi,(0<=w<=100000, 0<=x, y<=n) indicate the weight of node i, the left child of node i, and the right child of node i. If xi or yi equals to 0, it means node i has no such child.

And the following m lines describe the operations. There are two numbers p x in each line. If p is 0, it means do right-rotation on node x; If p is 1, it means do left-rotation on node x; if p is 2, it means ask the product of interesting values of the nodes in a subtree rooted at node x.

 

Output

For the k-th test case, first output “Case #k:” in a separate line.Then for each query,output a single num in each line after mod 1000000007.

 

Sample Input

13 41 2 31 0 01 0 02 10 12 22 1

 

Sample Output

Case #1:362
Hint
As the input data may cause the stack overflow if you use some recursion solution.Add #pragma comment (linker,"/STACK:102400000,102400000") in C++ language.

题意：给你一棵二叉树，支持左旋右旋操作，然后查询一个点的有趣值（子树节点权值的乘积）

思路：由于只支持左旋右旋，可以注意到这两种操作是不会改变整棵树的中序遍历的，由于每次旋转只会改变两个节点的信息，故可以直接用数组维护各个节点的信息，包括左

右孩子和父亲（我就是因为在旋转以后只换了孩子而没有换父亲就无限WA了 sad~），而每个节点的子树都是整棵树中序序列上的一段连续的区间，故可以用线段树点更新来

维护一段区间的乘积，突然觉得这题挺简单的= =。。