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?
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≤109
1≤Fi≤n, the root of the tree is node 1.
1≤u≤n,0≤x≤109
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≤109
1≤Fi≤n, the root of the tree is node 1.
1≤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：有一棵由n个节点构成的树，每个点上有一个权值，现在q次询问，每次输入u，x 表示以u为根节点的子树上 以某个节点上的权值异或x得到的最大值？
思路：持久化 trie 树，感觉和主席树差不多，是有 n 个版本的字典树，在遍历树的过程中经过点时，建立新的字典树，但实际上与旧的相比每次只是增加了log2（1e9）个节点，另外是在子树 u 上求最大异或值，所以需要保存 u 子树之前节点号和 u 子树中的最后一个节点的编号，作差即可得到相应的数据；
代码如下：
#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 ",query(root[la[u]-1],root[to[u]],0,x,30)); } } return 0; }