链接：http://codeforces.com/problemset/problem/570/D
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.
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.
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".
题意：
告诉你一颗树的父子关系，1节点为根。再告诉你每一个点上的字母。
问 v节点 子树（包含v节点）在第h行的全部节点的字母是否能组成回文串。
做法：
先用dfs 搜索 把全部节点标个左标号和右标号。 这样标号以后。每一个节点 用左标号 当自己 新的标号。 然后 子树全部节点 的新标号 肯定在 子树根节点的 左右标号之间。
标号之后分层来做。
每层 对每一个字母分别做统计。
把该层全部节点 的 左标号 在树状数组中+1. 然后对于该层的全部询问 做 树状数组统计。(sum(rit[v])-sum(lft[v]-1))。
假设是奇数 说明这个 字母在查询的区间内 有奇数个。
每一个查询 最多有一个奇数个的字母。否则不能构成回文串
#include <iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<vector>
#include<math.h>
using namespace std;
const int N = 500100;
int f[N];
vector<int> son[N];
int id;
int lft[N],rit[N];
int deps[N];
int ans[N];
char str[N];
vector<int> G[N];//深度
vector<pair<int,int> > Q[N];
void dfs(int nw,int dep)
{
lft[nw]=id++;
deps[nw]=dep;
G[dep].push_back(nw);
for(int i=0;i<son[nw].size();i++)
{
int to=son[nw][i];
dfs(to,dep+1);
}
rit[nw]=id++;
}
int bit[2*N];
int lowbit(int x)
{
return x&(-x);
}
void add(int wei,int x)
{
while(wei<=id)
{
bit[wei]+=x;
wei+=lowbit(wei);
}
}
int sum(int wei)
{
if(wei==0)
return 0;
int sum=0;
while(wei>0)
{
sum+=bit[wei];
wei-=lowbit(wei);
}
return sum;
}
int main()
{
int n,m;
while(scanf("%d%d",&n,&m)!=EOF)
{
for(int i=2;i<=n;i++)
{
scanf("%d",f+i);
son[f[i]].push_back(i);
}
scanf("%s",str+1);
id=1;
dfs(1,1);
int dep=1;
for(int i=1;i<=m;i++)
{
int vv,hh;
scanf("%d%d",&vv,&hh);
dep=max(hh,dep);
Q[hh].push_back(make_pair<int,int>(vv,i));
}
for(int i=1;i<=dep;i++)
{
for(int j=0;j<26;j++)
{
if(j==25)
int kkk=1;
for(int k=0;k<G[i].size();k++)//每一个节点
{
if(str[G[i][k]]-'a'==j)
add(lft[G[i][k]],1);
}
for(int k=0;k<Q[i].size();k++)
{
int v=Q[i][k].first;
int ii=Q[i][k].second;
if((sum(rit[v])-sum(lft[v]-1))&1) ans[ii]++;
}
for(int k=0;k<G[i].size();k++)//每一个节点
{
if(str[G[i][k]]-'a'==j)
add(lft[G[i][k]],-1);
}
// printf("jj %d
",j);
}
//printf("ii %d
",i);
}
for(int i=1;i<=m;i++)
{
if(ans[i]<=1)
printf("Yes
");
else
printf("No
");
}
}
return 0;
}