[Codeforces Round #316] Tree Requests

9
3
2015

Roman planted a tree consisting of n vertices. Each vertex contains a lowercase English letter. Vertex 1 is the root of the tree, each of the n - 1 remaining vertices has a parent in the tree. Vertex is connected with its parent by an edge. The parent of vertex i is vertex pi, the parent index is always less than the index of the vertex (i.e., pi < i).

The depth of the vertex is the number of nodes on the path from the root to v along the edges. In particular, the depth of the root is equal to 1.

We say that vertex u is in the subtree of vertex v, if we can get from u to v, moving from the vertex to the parent. In particular, vertex v is in its subtree.

Roma gives you m queries, the i-th of which consists of two numbers vi, hi. Let's consider the vertices in the subtree vi located at depthhi. Determine whether you can use the letters written at these vertices to make a string that is a palindrome. The letters that are written in the vertexes, can be rearranged in any order to make a palindrome, but all letters should be used.

Input

The first line contains two integers n, m (1 ≤ n, m ≤ 500 000) — the number of nodes in the tree and queries, respectively.

The following line contains n - 1 integers p2, p3, ..., pn — the parents of vertices from the second to the n-th (1 ≤ pi < i).

The next line contains n lowercase English letters, the i-th of these letters is written on vertex i.

Next m lines describe the queries, the i-th line contains two numbers vi, hi (1 ≤ vi, hi ≤ n) — the vertex and the depth that appear in thei-th query.

Output

Print m lines. In the i-th line print "Yes" (without the quotes), if in the i-th query you can make a palindrome from the letters written on the vertices, otherwise print "No" (without the quotes).

							Sample test(s)

									input
6 5
1 1 1 3 3
zacccd
1 1
3 3
4 1
6 1
1 2

									output
Yes
No
Yes
Yes
Yes

Note

String s is a palindrome if reads the same from left to right and from right to left. In particular, an empty string is a palindrome.

Clarification for the sample test.

In the first query there exists only a vertex 1 satisfying all the conditions, we can form a palindrome "z".

In the second query vertices 5 and 6 satisfy condititions, they contain letters "с" and "d" respectively. It is impossible to form a palindrome of them.

In the third query there exist no vertices at depth 1 and in subtree of 4. We may form an empty palindrome.

In the fourth query there exist no vertices in subtree of 6 at depth 1. We may form an empty palindrome.

In the fifth query there vertices 2, 3 and 4 satisfying all conditions above, they contain letters "a", "c" and "c". We may form a palindrome "cac".

题目大意：给出一棵有n个节点的树，每个节点对应一个字母。有m个询问，对于第i个询问，输出深度为$h_i$且属于以节点$v_i$为根的子树的所有节点所代表的所有字母在进行重新排列后能否构成一个回文串。

以下面这棵树为例，括号中的为每个节点代表的字母。

1(z)

|

---------------------

| | |

2(a) 3(c) 4(c)

| |

--------- ---------

| | | |

7(c) 8(d) 5(e) 6(e)

对树进行DFS，并记录每个深度下的所有节点（按DFS序排列）。下面就是记录的结果，括号中的为该节点的DFS序号。

深度为1: 1(1)

深度为2: 2(2), 3(3), 4(6)

深度为3: 7(4), 8(5), 5(7), 6(8)

对于询问i，在$D[h_i]$中用节点$v_i$的第一个和最后一个儿子的DFS序号进行二分查找，就可以获得询问要求的节点序列。

统计这些节点中各个字母的出现次数，若出现奇数次的字母不超过一个，就可以构成回文串。

暴力统计显然太慢。因为字母只有26个，所以可以用一个整数的各个二进制位来表示各个字母出现次数的奇偶性，DFS时顺便对每个深度的节点序列所代表的字母做前缀异或和，位运算乱搞就行了。

代码在此。

Category: 题解 | Tags: Codeforces 二分位运算树 | Read Count: 598

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