CF1900C.Anji's Binary Tree

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题目描述

Keksic keeps getting left on seen by Anji. Through a mutual friend, he's figured out that Anji really likes binary trees and decided to solve her problem in order to get her attention.

Anji has given Keksic a binary tree with nn vertices. Vertex 11 is the root and does not have a parent. All other vertices have exactly one parent. Each vertex can have up to 22 children, a left child, and a right child. For each vertex, Anji tells Keksic index of both its left and its right child or tells him that they do not exist.

Additionally, each of the vertices has a letter sis_i on it, which is either 'U', 'L' or 'R'.

Keksic begins his journey on the root, and in each move he does the following:

  • If the letter on his current vertex is 'U', he moves to its parent. If it doesn't exist, he does nothing.
  • If the letter on his current vertex is 'L', he moves to its left child. If it doesn't exist, he does nothing.
  • If the letter on his current vertex is 'R', he moves to its right child. If it doesn't exist, he does nothing.

Before his journey, he can perform the following operations: choose any node, and replace the letter written on it with another one. You are interested in the minimal number of operations he needs to do before his journey, such that when he starts his journey, he will reach a leaf at some point. A leaf is a vertex that has no children. It does not matter which leaf he reaches. Note that it does not matter whether he will stay in the leaf, he just needs to move to it. Additionally, note that it does not matter how many times he needs to move before reaching a leaf.

Help Keksic solve Anji's tree so that he can win her heart, and make her come to Čačak.

输入格式

Each test consists of multiple test cases. The first line contains a single integer tt ( 1t51041 \le t \le 5 \cdot 10^4 ) — the number of test cases. The description of test cases follows.

The first line of each test case contains a single integer nn ( 2n31052 \le n \le 3 \cdot 10^5 ) — the number of vertices in a tree.

The second line of each test case contains a string ss of nn characters — characters are written on the vertices. It is guaranteed that ss consists only of characters 'U', 'L', and 'R'.

The ii -th of the next nn lines contains two integers lil_i and rir_i ( 0li,rin0 \le l_i, r_i \le n ) — indices of left and right child of the vertex ii . If li=0l_i = 0 , it means that vertex ii does not have a left child. If ri=0r_i = 0 , it means that vertex ii does not have a right child. It is guaranteed that this data describes a valid binary tree rooted at 11 .

It is guaranteed that the sum of nn over all test cases does not exceed 31053 \cdot 10^5 .

输出格式

For each test case, output a single integer — the minimal number of operations Keksic needs to do to reach a leaf.

输入输出样例

  • 输入#1

    5
    3
    LRU
    2 3
    0 0
    0 0
    3
    ULR
    3 2
    0 0
    0 0
    2
    LU
    0 2
    0 0
    4
    RULR
    3 0
    0 0
    0 4
    2 0
    7
    LLRRRLU
    5 2
    3 6
    0 0
    7 0
    4 0
    0 0
    0 0

    输出#1

    0
    1
    1
    3
    1

说明/提示

In the first test case, vertex 11 has 22 as its left child and 33 as its right child. Vertices 22 and 33 do not have children and are therefore leaves. As 'L' is written on vertex 11 , Keksic will go to vertex 22 , therefore he has to do no operations.

In the second test case, vertex 11 has 33 as its left child and 22 as its right child. Vertices 22 and 33 are leaves. As 'U' is written on vertex 11 , Keksic needs to change it to either 'L' or 'R' in order for him to reach a leaf.

In the third case, vertex 11 has only a right child, which is vertex 22 . As 'L' is written on it, Keksic needs to change it to 'R', otherwise he would be stuck on vertex 11 .

In the fourth case, he can change 33 characters so that letters on the vertices are "LURL", which makes him reach vertex 22 .

In the fifth case, there are 33 leaves, 33 , 66 and 77 . To reach either leaf 66 or leaf 77 , he needs to change 22 characters. However, if he changes character on vertex 11 to 'R', he will reach leaf 33 , therefore the answer is 11 .

The initial tree in test case 5.

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