# hdu 6191-Query on A Tree（持久化字典树）

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hdu 6191--Query on A Tree（持久化字典树）

Problem Description
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

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

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <vector>
using namespace std;
const int N=1e5+5;
int a[N];
struct Node
{
int son[2];
int sum[2];
}node[35*N];
vector<int>G[N];
int la[N],to[N],root[N];
int tot1,tot2;

void init()
{
node[0].son[0]=node[0].son[1]=0;
node[0].sum[0]=node[0].sum[1]=0;
root[0]=0;
tot1=tot2=0;
for(int i=1;i<N;i++) G[i].clear();
}
void build(int pre,int now,int x,int deep)
{
if(deep<0) return ;
int tmp=!!(x&(1<<deep));
node[now]=node[pre];
node[now].sum[tmp]++;
build(node[pre].son[tmp],node[now].son[tmp]=++tot2,x,deep-1);
}
void dfs(int now)
{
la[now]=++tot1;
build(root[la[now]-1],root[la[now]]=++tot2,a[now],30);
for(int i=0;i<G[now].size();i++)
{
int v=G[now][i];
dfs(v);
}
to[now]=tot1;
}
int query(int pre,int now,int sum,int x,int deep)
{
if(deep<0) return sum;
int tmp=!!(x&(1<<deep));
if(node[now].sum[tmp^1]>node[pre].sum[tmp^1])
return query(node[pre].son[tmp^1],node[now].son[tmp^1],sum|(1<<deep),x,deep-1);
return query(node[pre].son[tmp],node[now].son[tmp],sum,x,deep-1);
}
int main()
{
int n,q;
while(scanf("%d%d",&n,&q)!=EOF)
{
init();
for(int i=1;i<=n;i++) scanf("%d",&a[i]);
for(int i=2;i<=n;i++)
{
int x; scanf("%d",&x);
G[x].push_back(i);
}
dfs(1);
while(q--)
{
int u,x; scanf("%d%d",&u,&x);
printf("%d\n",query(root[la[u]-1],root[to[u]],0,x,30));
}
}
return 0;
}