2017ACM/ICPC Guangxi Invitational K-query on A tree trie trees Merge

Reprinted from: http://www.cnblogs.com/zxhl/p/7459240.html

Problem Description
Monkey a lives on a tree, he is always plays on the 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 is curious about a problem.
The problem is what large the XOR result of number x and one node value of label Y can be, when giving you a non-negative i Nteger x and a node label u indicates that node y are in the subtree whose root are u (Y can be equal to u).
Can you help him?

Input
There is no more than 6 test cases.
For all test case there is positive integers n and q indicate that the tree have n nodes and you need to answer Q qu Eries.
Then the 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 is, integers u and x, indicating that node you pick should is in the subtree of U, and x H As 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

Problem:
Each number exists in the respective trie tree, n points This is a tree, and then from the bottom up the tri-Tree merged together
Query is to query a merged trie tree, using from high to low, greedy fetch

#include <bits/stdc++.h> inline int read () {int X=0,f=1;char ch=getchar (); while (ch< ' 0 ' | | Ch> ' 9 ') {if (ch== '-') F=-1;ch=getchar ();} while (ch>= ' 0 ' &&ch<= ' 9 ') {x=x*10+ch-' 0 '; Ch=getchar ();}

return x*f;}

using namespace Std;

#define LL Long Long const int N = 2E5;
Vector<int > G[n];
int n, q, x, U, a[n];

int ch[n*45][2], root[n],sz;
    void inserts (int u,int x) {Root[u] = ++sz;
    int tmp = SZ;
    int y = sz;
        for (int i = +; I >= 0; i) {int tmps = (x>>i) &1;
        if (!ch[y][tmps]) ch[y][tmps] = ++sz;
    y = Ch[y][tmps];
    }} int merges (int u,int to) {if (U = = 0) return to;
    if (to = = 0) return u;
    int t = ++sz;
    Ch[t][0] = merges (ch[u][0],ch[to][0]);
    CH[T][1] = merges (ch[u][1],ch[to][1]);
return t;
    } void Dfs (int u) {inserts (U,a[u]);
       for (auto To:g[u]) {DFS (to);
    Root[u] = merges (Root[u],root[to]);
    }} LL query (int u,int x) {int y = Root[u];
    LL ret = 0; for (int i= 30; I >= 0;
        -i) {int tmps = (x>>i) &1;
        if (ch[y][tmps^1]) ret + = (1<<i), y = ch[y][tmps^1];
    Else y = ch[y][tmps];
} return ret;
    } void Init () {for (int i = 0; I <= N; ++i) Root[i] = 0,g[i].clear ();
    SZ = 0;
memset (ch,0,sizeof (ch)); } int main (int argc, char * argv[]) {while (scanf ("%d%d", &n,&q)!=eof) {for (int i = 1; I <= N; + +)
        i) scanf ("%d", &a[i]);
        Init ();
            for (int i = 2; I <= n; ++i) {scanf ("%d", &x);
        G[x].push_back (i);
        } dfs (1);
            for (int i = 1; I <= Q; ++i) {scanf ("%d%d", &u,&x);
        printf ("%lld\n", Query (u,x));
}} return 0; }