HDU6191-Query on A Tree

Query on A Tree

                                                                   Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 132768/132768 K (Java/Others)
                                                                                                 Total Submission(s): 475    Accepted Submission(s): 182

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 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 21 211 32 1

 

Sample Output

23

 

Source

2017ACM/ICPC广西邀请赛-重现赛（感谢广西大学）

 

题意：给你一棵以1为根的树，每个节点有一个值，有q次询问，问x和以u为根节点子树中的一个节点的值异或后最大为多少

解题思路：字典树+启发式合并（注意启发式合并不能在线询问）

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <cmath>#include <map>#include <cmath>#include <set>#include <stack>#include <queue>#include <vector>#include <bitset>#include <functional>using namespace std;#define LL long longconst int INF = 0x3f3f3f3f;int n, m, a[100009], b[35], fa[100009];int ss[100009], f[35 * 100009][2], tot;int cnt, s[100009], nt[100009], e[100009];vector<pair<int, int> >g[100009];int ans[100009];void merge(int &now, int pre){    if (!now || !pre) { now = now^pre; return; }    merge(f[now][0], f[pre][0]);    merge(f[now][1], f[pre][1]);}void dfs(int k){    for (int i = s[k]; ~i; i = nt[i])    {        dfs(e[i]);        merge(ss[k], ss[e[i]]);    }    int Size = g[k].size();    for (int i = 0; i < Size; i++)    {        int x = g[k][i].first;        for (int i = 1; i <= 32; i++) b[i] = x % 2, x /= 2;        int p = ss[k], ans1 = 0;        for (int i = 32; i >= 1; i--)        {            ans1 <<= 1;            if (f[p][b[i] ^ 1]) ans1 |= 1, p = f[p][b[i] ^ 1];            else p = f[p][b[i]];        }        ans[g[k][i].second] = ans1;    }}int main(){    while (~scanf("%d%d", &n, &m))    {        tot = cnt = 0;        memset(s, -1, sizeof s);        for (int i = 1; i <= n; i++) scanf("%d", &a[i]), g[i].clear();        for (int i = 2; i <= n; i++)        {            scanf("%d", &fa[i]);            nt[cnt] = s[fa[i]], s[fa[i]] = cnt, e[cnt++] = i;        }        for (int i = 1; i <= n; i++)        {            for (int j = 1; j <= 32; j++) b[j] = a[i] % 2, a[i] /= 2;            ss[i] = ++tot;            memset(f[tot], 0, sizeof f[tot]);            for (int j = 32, p = tot; j >= 1; j--)            {                int k = b[j];                if (!f[p][k])                {                    f[p][k] = ++tot;                    memset(f[tot], 0, sizeof f[tot]);                }                p = f[p][k];            }        }        for (int i = 1; i <= m; i++)        {            int u, x;            scanf("%d%d", &u, &x);            g[u].push_back(make_pair(x, i));        }        dfs(1);        for (int i = 1; i <= m; i++) printf("%d\n", ans[i]);    }    return 0;}