[Codeforces 570D] Tree Requests dfs 시퀀스 + 2점

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).
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”. 제목:
n개의 점을 지정한 트리, m개의 질문 in[maxn],out[maxn];각 점마다 하나의 알파벳 질문형이 있는데 예를 들어 (u,deep) 질문 u점의 서브트리에서 뿌리의 깊이가 deep인 모든 점의 알파벳이 임의로 배열된 후에 회문열을 구성할 수 있는지, Yes.
사고방식: dfs 순서, 시간으로vectorsta[maxn][27]를 건설한다.
뿌리 깊이가 같은 점을vector에 저장합니다 D[i]는 깊이가 i인 모든 점을 표시하고 dfs에서 구할 수 있습니다.
질문을 깊이에 따라 정렬하고query[i]는 모든 깊이가 i인 질문을 표시합니다.
다음은 깊이에 따라 층층이 처리한다.
i층에 대해 i층에 있는 모든 노드를 26개의 트리 그룹으로 업데이트합니다.
그 다음에 질문을 처리하고 트리 수조에 몇 가지 자모가 홀수인지 직접 조회한다. 분명히 홀수 자모의 종수는 <=1
i층을 처리하면 트리 그룹을 역방향으로 조작하여 트리 그룹을 비우는 것과 같다
주의해야 할 점은 질문의 깊이가 임의라는 것이다. 즉, 실제 나무의 깊이를 초과할 수도 있고 현재 점의 깊이보다 작을 수도 있다.그래서 답을 초기화해야 하는데..

#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<vector>
#define maxn 500005
using namespace std;
vector<int> sta[maxn][27];
int in[maxn],out[maxn];
int time,n,m;
vector<int> lin[maxn];
int val[maxn];
void dfs(int x,int id)
{
    in[x]=++time;
    sta[id][val[x]].push_back(time);
    for(int i=0;i<lin[x].size();i++)  
        dfs(lin[x][i],id+1);
    out[x]=time;
}
//bool solve(int x,int dep)
//{
// int sum=0;
// for(int i=0;i<26;i++)
// {
// int num=(upper_bound(sta[dep][i].begin(),sta[dep][i].end(),out[x])-lower_bound(sta[dep][i].begin(),sta[dep][i].end(),in[x]));
// if(num%2)
// {
// sum++;
// if(sum>1) return false;
// }
// }
// return true;
//}
int main()
{
    scanf("%d%d",&n,&m);
    for(int i=2;i<=n;i++)
    {
        int aa;
        scanf("%d",&aa);
        lin[aa].push_back(i);
    }
    for(int i=1;i<=n;i++)
    {
        char c;
        scanf(" %c",&c);    
        val[i]=c-'a';
    }   
    dfs(1,1);
    while(m--)
    {
        int x;int dep;
        scanf("%d%d",&x,&dep);
        int sum=0;
        int inn=in[x];
        int ou=out[x];
        for(int i=0;i<26;i++)
        {
            int num=(upper_bound(sta[dep][i].begin(),sta[dep][i].end(),ou)-lower_bound(sta[dep][i].begin(),sta[dep][i].end(),inn));
            if(num%2)
            {
                sum++;
                if(sum>1) break;;
            }
        }
        if(sum>1) printf("No
");
        else
        printf("Yes
");
    }
}

비트 연산 최적화 후

#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<vector>
#define maxn 500005
using namespace std;
vector<int> sta[maxn];
int in[maxn],out[maxn];
int time,n,m;
int ma;
vector<int>f[maxn];
vector<int> lin[maxn];
int hash[maxn];
int val[maxn];
void dfs(int x,int id)
{
    ma=max(ma,id);
    in[x]=++time;
    hash[time]=val[x];
    f[id].push_back(time);
    for(int i=0;i<lin[x].size();i++)  
        dfs(lin[x][i],id+1);
    out[x]=time;
}
int main()
{
    scanf("%d%d",&n,&m);
    for(int i=2;i<=n;i++)
    {
        int aa;
        scanf("%d",&aa);
        lin[aa].push_back(i);
    }
    for(int i=1;i<=n;i++)
    {
        char c;
        scanf(" %c",&c);    
        val[i]=c-'a';
    }   
    dfs(1,1);
    for(int i=1;i<=ma;i++)  
    {  
        sta[i].push_back(1<<(hash[f[i][0]]));  
        for(int j=1;j<f[i].size();j++)  
        {  
            sta[i].push_back((1<<(hash[f[i][j]]))^sta[i][j-1]);  
        }  
    }  
    while(m--)
    {
        int x;int dep;
        scanf("%d%d",&x,&dep);
        int sum=0;
        int l =lower_bound(f[dep].begin(),f[dep].end(),in[x])-f[dep].begin();  
        int r =upper_bound(f[dep].begin(),f[dep].end(),out[x])-f[dep].begin();

        l--;
        r--;  
        if(r<0)     printf("Yes
");
        else
        {
                int tt=sta[dep][r];
        if(l>=0)
        tt^=sta[dep][l];
        sum=(__builtin_popcount(tt)); 
        if(sum>1) printf("No
");
        else
        printf("Yes
");
        }

    }
}