stack ott ise ger query numbers mem 1.3 locate

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題意：

給定n個點的樹。m個詢問

以下n-1個數給出每一個點的父節點，1是root

每一個點有一個字母

以下n個小寫字母給出每一個點的字母。

以下m行給出詢問：

詢問形如 (u, deep) 問u點的子樹中，距離根的深度為deep的全部點的字母是否能在隨意排列後組成回文串，能輸出Yes.

思路：dfs序，給點又一次標號，dfs進入u點的時間戳記為l[u], 離開的時間戳記為r[u], 這樣對於某個點u，他的子樹節點相應區間都在區間 [l[u], r[u]]內。

把距離根深度同樣的點都存到vector裏 D[i] 表示深度為i的全部點，在dfs時能夠順便求出。

把詢問按深度排序,query[i]表示全部深度為i的詢問。

接下來依照深度一層層處理。

對於第i層，把全部處於第i層的節點都更新到26個樹狀數組上。

然後處理詢問，直接查詢樹狀數組上有多少種字母是奇數個的。顯然奇數個字母的種數要<=1

處理完第i層，就把樹狀數組逆向操作。相當於清空樹狀數組

註意的一個地方就是 詢問的深度是隨意的，也就是說可能超過實際樹的深度，也可能比當前點的深度小。

。

所以須要初始化一下答案。。

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <stdio.h>  
#include <iostream>  
#include <algorithm>  
#include <sstream>  
#include <stdlib.h>  
#include <string.h>  
#include <limits.h>  
#include <vector>  
#include <string>  
#include <time.h>  
#include <math.h>  
#include <iomanip>  
#include <queue>  
#include <stack>  
#include <set>  
#include <map>  
const int inf = 1e9;
const double eps = 1e-8;
const double pi = acos(-1.0);
template <class T>
inline bool rd(T &ret) {
	char c; int sgn;
	if (c = getchar(), c == EOF) return 0;
	while (c != ‘-‘ && (c<‘0‘ || c>‘9‘)) c = getchar();
	sgn = (c == ‘-‘) ? -1 : 1;
	ret = (c == ‘-‘) ? 0 : (c - ‘0‘);
	while (c = getchar(), c >= ‘0‘&&c <= ‘9‘) ret = ret * 10 + (c - ‘0‘);
	ret *= sgn;
	return 1;
}
template <class T>
inline void pt(T x) {
	if (x < 0) { putchar(‘-‘); x = -x; }
	if (x > 9) pt(x / 10);
	putchar(x % 10 + ‘0‘);
}
using namespace std;
const int N = 5e5 + 100;
typedef long long ll;
typedef pair<int, int> pii;
struct BIT {
	int c[N], maxn;
	void init(int n) { maxn = n; memset(c, 0, sizeof c); }
	inline int Lowbit(int x) { return x&(-x); }
	void change(int i, int x)//i點增量為x
	{
		while (i <= maxn)
		{
			c[i] += x;
			i += Lowbit(i);
		}
	}
	int sum(int x) {//區間求和 [1,x]
		int ans = 0;
		for (int i = x; i >= 1; i -= Lowbit(i))
			ans += c[i];
		return ans;
	}
	int query(int l, int r) {
		return sum(r) + sum(l - 1); 
	}
}t[26];
int n, m;
char s[N];
vector<int>G[N], D[N];
int l[N], r[N], top;
vector<pii>query[N];
bool ans[N];
void dfs(int u, int fa, int dep) {
	D[dep].push_back(u);
	l[u] = ++top;
	for (auto v : G[u])
		if (v != fa)dfs(v, u, dep + 1);
	r[u] = top;
}
int main() {
	rd(n); rd(m);
	fill(ans, ans + m + 10, 1);
	for (int i = 0; i < 26; i++) t[i].init(n);
	for (int i = 2, u; i <= n; i++)rd(u), G[u].push_back(i);
	top = 0;
	dfs(1, 1, 1);
	scanf("%s", s + 1);
	for (int i = 1, u, v; i <= m; i++) {
		rd(u); rd(v); query[v].push_back(pii(u, i));
	}
	for (int i = 1; i <= n; i++)
	{
		if (D[i].size() == 0)break;
		for (auto v : D[i])	t[s[v] - ‘a‘].change(l[v], 1);
		
		for (pii Q : query[i])
		{
			int ou = 0;
			for (int j = 0; j < 26; j++)
			{
				if (t[j].query(l[Q.first], r[Q.first]))
					ou += t[j].query(l[Q.first], r[Q.first]) & 1;
			}
			ans[Q.second] = ou <= 1;
		}
		for (auto v : D[i])	t[s[v] - ‘a‘].change(l[v], -1);
	}
	for (int i = 1; i <= m; i++)ans[i] ?
 
 puts("Yes") : puts("No");

	return 0;
}

D. Tree Requests time limit per test 2 seconds memory limit per test 256 megabytes input standard input output 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.

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 thei-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".

Codeforces 570D TREE REQUESTS dfs序+樹狀數組