D. Tree Requests
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard 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 vi, hi. Let’s consider the vertices in the subtree vi located at depth hi. 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 the i-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).
Examples
inputCopy
6 5
1 1 1 3 3
zacccd
1 1
3 3
4 1
6 1
1 2
outputCopy
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”.
dfs序学自：https://blog.csdn.net/u012061345/article/details/54023285
简单来说，就是记录dfs中每一个节点入栈和出栈的时间，如果a节点入栈时间在b节点入栈出栈时间之间的话，a即为b子树上的节点。
用一个二维数组存储每个深度所有的节点的到达时间，并用二进制数记录这个节点是什么字母，搜索时先锁定深度，在从中找到v的子节点，将这些节点进行异货，一串字符能组成回文串当且仅当最多有一个字母出现次数为奇数，因此判断异货后的数中1的个数即可。
/* ***********************************************
Author :ACagain
Created Time :2018/4/10 19:53:20
File Name :4_10.cpp
************************************************ */
#include <cstdio>
#include <algorithm>
#include <iostream>
#include <cmath>
#include <vector>
using namespace std;
#define lson o<<1,l,m
#define rson o<<1|1,m+1,r
#define pii pair<int,int>
#define mp make_pair
#define ll long long
#define INF 0x3f3f3f3f
const int maxn=5*1e5+2;
int n,m,t;
char c[maxn];
vector<int> son[maxn];
vector<int> d[maxn]; //d[i][j] i深度第j个节点到达时间
int l[maxn],r[maxn]; //v入栈出栈时间
vector<int> hsh[maxn]; //对应字母
void dfs(int x,int depth)
{
l[x]=++t;
int k=son[x].size();
int tmp=1<<(int(c[x]-'a'));
d[depth].push_back(t);
int tt=hsh[depth].size();
if(tt==0)
hsh[depth].push_back(tmp);
else
hsh[depth].push_back(tmp^hsh[depth][tt-1]);
for(int i=0;i<k;i++)
dfs(son[x][i],depth+1);
r[x]=t;
}
int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
scanf("%d%d",&n,&m);
int tmp;
for(int i=2;i<=n;i++)
{
scanf("%d",&tmp);
son[tmp].push_back(i);
}
for(int i=1;i<=n;i++)
scanf(" %c",&c[i]);
t=-1;
dfs(1,1);
int v,h,num;
while(m--)
{
scanf("%d%d",&v,&h);
num=0;
if(d[h].size()==0)
{
printf("%s
","Yes");
continue;
}
int L=lower_bound(d[h].begin(),d[h].end(),l[v])-d[h].begin()-1;
int R=upper_bound(d[h].begin(),d[h].end(),r[v])-d[h].begin()-1;
int tmp=hsh[h][R];
if(R<0)
{
printf("%s
","Yes");
continue;
}
if(L>=0)
tmp^=hsh[h][L];
while(tmp)
{
if(tmp&1==1)
num++;
tmp>>=1;
}
if(num>1)
printf("%s
","No");
else
printf("%s
","Yes");
}
return 0;
}