[DFS序上建可持久化字典树贪心异或最大值] Query on A Tree HDU

题意：给出一颗树和Q个询问

对于每个询问 x y

你需要搞出 x 节点 子树上一点 (包括自己) 与 y 值异或的最大值

 解题思路：

套路题

子树 → dfs序

异或极值 → trie树

dfs + trie → 可持久化trie

按连续序列建树即可

代码

---

/*
    Zeolim - An AC a day keeps the bug away
*/
 
//#pragma GCC optimize(2)
//#pragma GCC ("-W1,--stack=128000000")
#include <bits/stdc++.h>
using namespace std;
#define mp(x, y) make_pair(x, y)
#define fr(x, y, z) for(int x = y; x < z; ++x)
#define pb(x) push_back(x)
#define mem(x) memset(x, 0, sizeof(x))
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef std::pair <int, int> pii;
typedef std::vector <int> vi;
//typedef __int128 ill;
const ld PI = acos(-1.0);
const ld E = exp(1.0);
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll MOD = 386910137;
const ull P = 13331; 
const int MAXN = 2e5 + 10;

struct trie
{
	int cnt, root[MAXN], tr[MAXN * 30][2], sum[MAXN * 30];
	trie() {cnt = 0;}
	void insert(int now, int last, int val) 
	{
		int nto, lto;
		root[now] = nto = ++cnt;
		lto = root[last];
		for(int i = 30; i >= 0; --i)
		{
			int dig = val >> i & 1;
			tr[nto][0] = tr[lto][0];
			tr[nto][1] = tr[lto][1];  		 //½«ÉÏÒ»°æ±¾¶ù×ÓÖ±½Ó¸²¸Ç 
			tr[nto][dig] = ++cnt, nto = cnt; //µ±Ç°°æ±¾µÄжù×Ó 
			lto = tr[lto][dig]; 
			sum[nto] = sum[lto] + 1; 		 //ǰ׺ºÍ + 1 
		}
	}
	int get(int now, int last, int val)
	{
		int ret = 0;
		int nto = root[now], lto = root[last];
		for(int i = 30; i >= 0; --i)
		{
			int dig = val >> i & 1 ^ 1;
			if(sum[ tr[nto][dig] ] - sum[ tr[lto][dig] ]) //µ±Ç°Çø¼äÄÚÓиÃÄ¿±ê½Úµã 
				ret += (1 << i), nto = tr[nto][dig], lto = tr[lto][dig];
			else
				nto = tr[nto][dig ^ 1], lto = tr[lto][dig ^ 1];
		}
		return ret;
	}
	void clear()
	{
		cnt = 0;
		mem(sum);
		mem(tr);
		mem(root);
	}
}T;

int val[MAXN] = {0};

int fa[MAXN] = {0};

vector <int> edge[MAXN];

int arr[MAXN] = {0}, tot = 0;

int in[MAXN] = {0}, out[MAXN] = {0};

void gao(int now)
{
	arr[++tot] = val[now];
	in[now] = tot;
	for(auto x :edge[now])
		gao(x);
	out[now] = tot;
}

void clear(int n)
{
	T.clear();
	for(int i = 0; i <= n; ++i)
		edge[i].clear();
	tot = 0;
}

int main()
{  
	ios::sync_with_stdio(0);
	cin.tie(0); cout.tie(0);
	
	int n, q;
	
	while(cin >> n >> q)
	{
		clear(n);
		
		for(int i = 1; i <= n; ++i)
			cin >> val[i];
	
		for(int i = 2; i <= n; ++i)
		{
			cin >> fa[i];
			edge[fa[i]].push_back(i);
		}
		
		gao(1);
		
		T.insert(0, 0, 0);
		
		for(int i = 1; i <= n; ++i)
		{
			T.insert(i, i - 1, arr[i]);
		}
		
		int x, y;
		
		while(q--)
		{
			cin >> x >> y;
			
			cout << T.get(out[x], in[x] - 1, y) << '
';
		}
	}
	
	
	
    return 0;
}

/*
5 5
5 2 7 3 1
3 1 1 3 
1 5
3 2
4 3
2 1
5 5
*/