HDU

Monkey A lives on a tree, he always plays on this tree.

One day, monkey A learned about one of the bit-operations, xor. He was keen of this interesting operation and wanted to practise it at once.

Monkey A gave a value to each node on the tree. And he was curious about a problem.

The problem is how large the xor result of number x and one node value of label y can be, when giving you a non-negative integer x and a node label u indicates that node y is in the subtree whose root is u(y can be equal to u).

Can you help him?

Input

There are no more than 6 test cases.

For each test case there are two positive integers n and q, indicate that the tree has n nodes and you need to answer q queries.

Then two lines follow.

The first line contains n non-negative integers V1,V2,⋯,Vn, indicating the value of node i.

The second line contains n-1 non-negative integers F1,F2,⋯Fn−1, Fi means the father of node i+1.

And then q lines follow.

In the i-th line, there are two integers u and x, indicating that the node you pick should be in the subtree of u, and x has been described in the problem.

2≤n,q≤105
0≤Vi≤10^9
1≤Fi≤n, the root of the tree is node 1.
1≤u≤n,0≤x≤10^9

Output

For each query, just print an integer in a line indicating the largest result.

Sample Input

2 2
1 2
1
1 3
2 1

Sample Output

2
3

题解：

01字典树就不多说了，不会的这题也没必要做了。DFS序就是对一个树跑一边DFS，遍历点的顺序就是DFS序。

主要是启发式合并，这题做法其实就是给每个点建立一个01字典树，而通过DFS可以保证每个点的01字典树中的点都是其自身或子节点，而启发式合并的作用就在于节省了建立01字典树的空间和时间。

ps：启发式合并时间复杂度o(nlogn*插入复杂度的）。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <vector>

using namespace std;

typedef struct Node* node;

const int MAXN = 1e5+10;

struct Edge{
	int to,next;
}E[MAXN*4];

int head[MAXN],tot;
int board[MAXN];//存点的值 
int re[MAXN];//存每个问题的结果 
vector< pair<int,int> >P[MAXN];//分别存以每个点为根节点的问题。 

void Add(int from,int to){
	E[++tot].next = head[from];
	head[from] = tot;
	E[tot].to = to;
}

struct Node
{
	int num,val;
	node Next[2];
	Node()
	{
		val = num = 0;
		memset(Next,NULL,sizeof(Next));
	}
}*root[MAXN];

void Insert(node root,int x,int flag)
{
	node p = root;
	for(int i=31 ; i>=0 ; i--)
	{
		int t = (x>>i)&1;
		if(p->Next[t] == NULL)p->Next[t] = new struct Node();
		p->num += flag;
		p = p->Next[t]; 
	}
	p->num += flag;
	p->val = x;
}

int Judge(node root,int x)
{
	node p = root;
	for(int i=31 ; i>=0 ; i--)
	{
		int t = ((x>>i)&1)^1;
		if(p->Next[t] == NULL || p->Next[t]->num == 0)t = (x>>i)&1;
		if(p->Next[t] && p->Next[t]->num)p = p->Next[t];
		else return 0;
	}
	return p->val;
} 

void Del(node root){
	for(int i=0 ; i<2 ; ++i){
		if(root->Next[i])Del(root->Next[i]);
	}
	delete(root);
}

inline void init(){
	memset(head,0,sizeof head);
	tot = 0;
}

node Merge(node p,node q){
	if(p == NULL)return q;
        else if(q == NULL)return p;
        else p->num += q->num;
	p->Next[0] = Merge(p->Next[0],q->Next[0]);
	p->Next[1] = Merge(p->Next[1],q->Next[1]);
	free(q);
	return p;
}

void DFS(int now,int father){
	root[now] = new struct Node();
	Insert(root[now],board[now],1);
	for(int i=head[now] ; i ; i=E[i].next){
		if(E[i].to == father)continue; 
		DFS(E[i].to,now);
		root[now] = Merge(root[now],root[E[i].to]);
	}
	for(int i=0 ; i<P[now].size() ; ++i){
		re[P[now][i].first] = P[now][i].second^Judge(root[now],P[now][i].second);
	}
}

int main(){
	
	int N,Q;
	while(scanf("%d %d",&N,&Q)!=EOF){
		init();
		for(int i=1 ; i<=N ; ++i){
			scanf("%d",&board[i]);
			P[i].clear();
		}
		for(int i=2 ; i<=N ; ++i){
			int t;
			scanf("%d",&t);
			Add(t,i);
		}
		for(int i=1 ; i<=Q ; ++i){
			int a,b;
			scanf("%d %d",&a,&b);
			P[a].push_back(make_pair(i,b));//pair.first代表问题进来的顺序 
		} 
		DFS(1,-1);
		for(int i=1 ; i<=Q ; ++i)printf("%d
",re[i]);
		Del(root[1]);
	}
	
	return 0;
}