2 seconds
256 megabytes
standard input
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 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：给你一棵树，n个节点，m个询问；根节点为1；根结点的深度为1，
每个节点含有一个权值；
询问给你一个节点node，以node为根的子树中深度为x的结点，能否形成回文串；
思路：首先，能行成回文，只需要奇数的字母个数<=1；状态压缩标记即可；
然后处理这课树，dfs序形成一个序列处理；
对于m个询问， 两种思路：
1：打表每个字母+深度进行存in[i]（vector），然后m个询问复杂度o(26*（m*log（n））；
2：可以再广搜一遍，对于深度相同的必然是连续的一段序列，利用前缀异或和处理，复杂度（m*log（n）);
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define pi (4*atan(1.0))
#define eps 1e-14
const int N=5e5+,M=1e6+,inf=1e9+;
const ll INF=1e18+,mod=;
int in[N],out[N],tot,a[N],flag[N],deep[N];
struct is
{
int v,nex;
}edge[N<<];
int head[N<<],edg;
int n,p;
void init()
{
memset(head,-,sizeof(head));
memset(a,,sizeof(a));
edg=;
tot=;
}
void add(int u,int v)
{
edg++;
edge[edg].v=v;
edge[edg].nex=head[u];
head[u]=edg;
}
void dfs(int u,int fa,int d)
{
deep[u]=d;
in[u]=++tot;
for(int i=head[u];i!=-;i=edge[i].nex)
{
int v=edge[i].v;
if(v==fa)continue;
dfs(v,u,d+);
}
out[u]=tot;
}
char ch[N];
int root,x;
vector<int>ans[][N];
int main()
{
int n,q;
while(~scanf("%d%d",&n,&q))
{
init();
for(int i=;i<=n;i++)
{
int v;
scanf("%d",&v);
add(i,v);
add(v,i);
}
dfs(,-,);
scanf("%s",ch+);
//for(int i=1;i<=n;i++)
// cout<<in[i]<<" "<<out[i]<<endl;
for(int i=;i<=n;i++)
flag[in[i]]=i;
for(int i=;i<=n;i++)
a[i]=deep[flag[i]];
for(int i=;i<=n;i++)
ans[ch[flag[i]]-'a'][a[i]].push_back(in[flag[i]]);
for(int i=;i<;i++)
{
for(int j=;j<=n;j++)
{
sort(ans[i][j].begin(),ans[i][j].end());
}
}
/*for(int i=0;i<26;i++)
{
for(int j=1;j<=n;j++)
{
cout<<j<<" ";
for(int k=0;k<ans[i][j].size();k++)
cout<<ans[i][j][k]<<" ";
cout<<endl;
}
cout<<"~~~~"<<endl;
}*/
while(q--)
{
scanf("%d%d",&root,&x);
if(deep[root]>=x)
{
printf("Yes\n");
continue;
}
int tot=;
for(int i=;i<;i++)
{
int pos1=lower_bound(ans[i][x].begin(),ans[i][x].end(),in[root])-ans[i][x].begin()-;
int pos2=upper_bound(ans[i][x].begin(),ans[i][x].end(),out[root])-ans[i][x].begin()-;
//cout<<pos1<<" "<<pos2<<endl;
tot+=(pos2-pos1)%;
}
if(tot>=)
printf("No\n");
else
printf("Yes\n");
}
}
return ;
}