【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,⋯,VnV1,V2,⋯,Vn, indicating the value of node i.

The second line contains n-1 non-negative integers F1,F2,⋯Fn−1F1,F2,⋯Fn−1, FiFi means the father of node i+1i+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≤1052≤n,q≤105

0≤Vi≤1090≤Vi≤109

1≤Fi≤n1≤Fi≤n, the root of the tree is node 1.

1≤u≤n,0≤x≤1091≤u≤n,0≤x≤109
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
题意 ：给出一棵树，树上每个点有点权，每次询问给出u x，求以u为根的子树中点权和x异或得到的最大值
分析： 可以先把这个题写了
然后把树用DFS序处理之后也就是，区间查询了。
代码

#include<bits/stdc++.h>using namespace std;typedef pair<int,int>pii;#define first fi#define second se#define  LL long long#define fread() freopen("in.txt","r",stdin)#define fwrite() freopen("out.txt","w",stdout)#define CLOSE() ios_base::sync_with_stdio(false)const int MAXN = 1e5+11;const int MAXM = 1e6;const int mod = 1e9+7;const int inf = 0x3f3f3f3f;int root[MAXN*35],sum[MAXN*35],son[MAXN*35][2],sz; void init(){    memset(root,0,sizeof(root));    memset(sum,0,sizeof(sum));    memset(son,0,sizeof(son));    sz=0;}int Insert(int val,int pre){    int x=++sz,t=x;    for(int i=31;i>=0;i--){        son[x][1]=son[pre][1],son[x][0]=son[pre][0];          sum[x]=sum[pre]+1;        int j= 1&(val>>i);        son[x][j]=++sz;           x=son[x][j],pre=son[pre][j];    }    sum[x]=sum[pre]+1;    return t;   } int Query(int val,int le,int ri){    int ans=0;    for(int i=31;i>=0;i--){        int j=1&(val>>i); j=!j;        if(sum[son[ri][j]]-sum[son[le][j]]>0) {            ans|=(1<<i);              le=son[le][j],ri=son[ri][j];        }else             le=son[le][!j],ri=son[ri][!j];    }    return ans;}int cost[MAXN],in[MAXN],out[MAXN],id;vector<int>ve[MAXN];void dfs(int now,int pre){ // 得到DFS序    in[now]=id++; root[in[now]]= Insert( cost[now], root[in[now]-1]);     for(int i=0;i<ve[now].size();i++){        if(ve[now][i]==pre) continue;        dfs(ve[now][i],now);    }    out[now]=id-1;}int main(){    int n,q;    while(scanf("%d%d",&n,&q)!=EOF){        memset(in,0,sizeof(in));        memset(out,0,sizeof(out));        init();        for(int i=1;i<=n;i++)  { scanf("%d",&cost[i]); ve[i].clear(); }        for(int i=1;i<=n-1;i++) {            int a;scanf("%d",&a);            ve[a].push_back(i+1);        }        id=1;  dfs(1,-1);//      for(int i=1;i<=n;i++) {//          printf("in %d   out %d \n",in[i],out[i]);//      }        while(q--){            int u,x;scanf("%d%d",&u,&x);            printf("%d\n", Query(x,root[in[u]-1],root[out[u]]));        }    }    return 0;}