Codeforces Round 316 (Div. 2) D题 Tree Requests（二分，dfs，在线，前缀异或）

题目链接

Codeforces Round 316 (Div. 2) D题 Tree Requests

思路

将 26 26 26个字母全部当作一个二进制数。

将每个深度的结点按照dfs序放到一个vector里，同时记录每个vector对应的前缀异或。

对于每一个询问x，只需在给定深度里找到 ≥ \ge ≥L[x]和 ≤ \le ≤R[x]的两个端点，取区间异或和即可。

代码

#include <bits/stdc++.h>
using namespace std;
#define int long long
const int N = 5e5 + 5;
int n, m;
int depth[N], L[N], R[N], ID[N], tim;
char s[N];
vector<int>mp[N], XOR[N], dep[N];
int lowbit(int x) {return (x) & (-x);}
void dfs(int u, int deep)
{
	tim++;
	depth[u] = deep;
	L[u] = tim;
	ID[tim] = u;
	dep[deep].push_back(L[u]);
	for (int j : mp[u])
	{
		dfs(j, deep + 1);
	}
	R[u] = tim;
}
void solve()
{
	cin >> n >> m;
	for (int i = 2, p; i <= n; i++)
	{
		cin >> p;
		mp[p].push_back(i);
	}
	for (int i = 1; i <= n; i++)
	{
		cin >> s[i];
	}
	dfs(1, 1);
	int maxdeep = *max_element(depth + 1, depth + 1 + n);
	for (int i = 1; i <= maxdeep; i++)
	{
		XOR[i].push_back(1ll << (s[ID[dep[i][0]]] - 'a'));
		for (int j = 1; j < dep[i].size(); j++)
		{
			XOR[i].push_back(XOR[i][j - 1] ^ (1ll << (s[ID[dep[i][j]]] - 'a')));
		}
	}

	while (m--)
	{
		int x, d, ans = 0;
		cin >> x >> d;
		int l = lower_bound(dep[d].begin(), dep[d].end(), L[x]) - dep[d].begin();
		int r = upper_bound(dep[d].begin(), dep[d].end(), R[x]) - dep[d].begin();
		r--;
		if (r < 0)
		{
			cout << "Yes" << endl;
		}
		else
		{
			if (l == 0) ans = XOR[d][r];
			else ans = XOR[d][l - 1] ^ XOR[d][r];
			if (ans == lowbit(ans))
			{
				cout << "Yes" << endl;
			}
			else cout << "No" << endl;
		}
	}
}
signed main()
{
	ios::sync_with_stdio(false);
	cin.tie(0), cout.tie(0);
	int test = 1;
	// cin >> test;
	for (int i = 1; i <= test; i++)
	{
		solve();
	}
	return 0;
}