Tree Requests-Codeforces Round #316 (Div. 2)

D. Tree Requests
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

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 vihi. 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 nm (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 vihi (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".

题意:给一棵树,树上各个节点对应一个字符,m个询问v,h,即v所在的子树节点中深度为h(在整棵树上的深度)的点的字符能否组成回文串。

思路:

Tree Requests-Codeforces Round #316 (Div. 2)_第1张图片

即dfs记录子树的出入时间,并将同一深度的同一字符保存在一起,对于输入的v以in[v]和out[v] 在对应的h深度不同字符段做二分查找,判断个数奇偶。

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
#define eps 1e-8
#define inf 0x3f3f3f3f
#define rd(x) scanf("%d",&x)
#define rd2(x,y) scanf("%d%d",&x,&y)
#define ll long long int
#define mod 1000000007
#define maxn 500005
#define maxm 300005
vector ch[maxn];
vector f[maxn][26];
int n,m,x,tot;
string s;
int in[maxn],out[maxn];
void dfs(int x,int k){
    in[x]=++tot;
    f[k][s[x-1]-'a'].push_back(tot);//加入保存该深度的该字符的vector
    for(int i=0;i>s;
    tot=0;
    dfs(1,1);
    for(int i=1;i<=m;i++)
    {
        rd2(v,h);
        tot=0;
        for(int i=0;i<26;i++)
        {
            int res=lower_bound(f[h][i].begin(),f[h][i].end(),out[v]+1)-
                     lower_bound(f[h][i].begin(),f[h][i].end(),in[v]);//二分查找满足要求的该字符的起止位置
                if(res%2) tot++;
                if(tot>1) break;
        }
        if(tot>1) printf("No\n");
        else printf("Yes\n");
    }
    return 0;
}


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