做法：
推断能否够构成一个回文串的话，仅仅须要知道是否有大于一个的奇数数目的字母就可以。为了非常快的訪问到一个区间。记录前缀和就可以。为了省内存，状压奇偶就可以。
为了非常快的找到以结点v为根的子树中高度为h的后代,须要dfs整棵树。然后记录每一个结点第一次訪问它的时间戳以及离开它的时间戳，就能够二分出来。
为了省内存，能够离线处理询问。
#include<map>
#include<string>
#include<cstring>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<queue>
#include<vector>
#include<iostream>
#include<algorithm>
#include<bitset>
#include<climits>
#include<list>
#include<iomanip>
#include<stack>
#include<set>
#include<bitset>
using namespace std;
int head[500010],tail;
struct Edge
{
int to,next;
}edge[500010];
void add(int from,int to)
{
edge[tail].to=to;
edge[tail].next=head[from];
head[from]=tail++;
}
int in[500010],ot[500010];
int cnt;
struct node
{
int id,tm;
node(){}
node(int id,int tm)
{
this->id=id;
this->tm=tm;
}
bool operator <(node a)const
{
return tm<a.tm;
}
};
vector<node>bx[500010];
void dfs(int from,int step)
{
in[from]=++cnt;
bx[step].push_back(node(from,cnt));
for(int i=head[from];i!=-1;i=edge[i].next)
{
int to=edge[i].to;
dfs(to,step+1);
}
ot[from]=++cnt;
}
bool num[500010][26];
char s[500010];
void create(int h)
{
int n=bx[h].size();
for(int i=0;i<n;i++)
{
if(i==0)
{
for(int j=0;j<26;j++)
num[i][j]=0;
}
else
{
for(int j=0;j<26;j++)
num[i][j]=num[i-1][j];
}
int id=bx[h][i].id-1;
num[i][s[id]-'a']^=1;
}
}
bool work(int v,int h)
{
if(bx[h].empty())
return 1;
int l=upper_bound(bx[h].begin(),bx[h].end(),node(-1,in[v]))-bx[h].begin();
if(l==bx[h].size()||bx[h][l].tm>ot[v])
return 1;
int r=lower_bound(bx[h].begin(),bx[h].end(),node(-1,ot[v]))-bx[h].begin();
l--;r--;
bool flag=0;
for(int i=0;i<26;i++)
{
bool t;
if(l<0)
t=num[r][i];
else
t=(num[l][i]^num[r][i]);
if(t&1)
{
if(flag)
return 0;
flag=1;
}
}
return 1;
}
struct Q
{
int id,v,h;
bool operator <(Q a)const
{
return h<a.h;
}
}q[500010];
bool ans[500010];
int main()
{
int n,m;
scanf("%d%d",&n,&m);
memset(head,-1,sizeof(head));
for(int i=2;i<=n;i++)
{
int p;
scanf("%d",&p);
add(p,i);
}
scanf("%s",s);
dfs(1,1);
for(int i=0;i<m;i++)
{
scanf("%d%d",&q[i].v,&q[i].h);
q[i].id=i;
}
sort(q,q+m);
int p=-1;
for(int i=0;i<m;i++)
{
if(q[i].h!=p)
{
p=q[i].h;
create(p);
}
ans[q[i].id]=work(q[i].v,q[i].h);
}
for(int i=0;i<m;i++)
if(ans[i])
puts("Yes");
else
puts("No");
}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 then - 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.
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.
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).
6 5 1 1 1 3 3 zacccd 1 1 3 3 4 1 6 1 1 2
Yes No Yes Yes Yes
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".