HDU6191(01字典树启发式合并)

Query on A Tree

Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 132768/132768 K (Java/Others)
Total Submission(s): 801    Accepted Submission(s): 302

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 2 1 2 1 1 3 2 1

 

Sample Output

2 3

 

Source

2017ACM/ICPC广西邀请赛-重现赛（感谢广西大学）

题意：给一棵树，树上每个节点有一个权值。然后有q次查询，每次查询给你节点标号u和一个数x。问以u的子树里面的所有节点点权和数x的最大异或值是多少。

思路：使用01字典树解决异或值最大问题，使用字典树的合并，可得到每棵子树上的异或最大值。

//2017-09-16
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>

using namespace std;

const int N = 110000;
int head[N], tot;
struct Edge{
    int v, next;
}edge[N];

void init(){
    tot = 0;
    memset(head, -1, sizeof(head));
}

void add_edge(int u, int v){
    edge[tot].v = v;
    edge[tot].next = head[u];
    head[u] = tot++;
}

int n, q, arr[N];
int ans[N];
vector< pair<int, int> > qy[N];
struct Trie{
    Trie* next[2];
}*root[N];

void insert(Trie* rt, int x){
    for(int i = 31; i >= 0; i--){
        int idx = (x>>i)&1;
        if(rt->next[idx] == NULL){
            Trie *tmp = new Trie();
            rt->next[idx] = tmp;
        }
        rt = rt->next[idx];
    }
}

int query(Trie* rt, int x){
    int ans = 0;
    for(int i = 31; i >= 0; i--){
        int idx = (x>>i)&1;
        if(rt->next[idx^1] != NULL){
            rt = rt->next[idx^1];
            ans |= (1<<i);
        }else rt = rt->next[idx];
    }
    return ans;
}

Trie* merge(Trie* p, Trie* q){
    if(p == NULL)return q;
    if(q == NULL)return p;
    p->next[0] = merge(p->next[0], q->next[0]);
    p->next[1] = merge(p->next[1], q->next[1]);
    free(q);
    return p;
}

void dfs(int u, int fa){
    root[u] = new Trie();
    insert(root[u], arr[u]);
    for(int i = head[u]; i != -1; i = edge[i].next){
        int v = edge[i].v;
        if(v == fa)continue;
        dfs(v, u);
        root[u] = merge(root[u], root[v]);
    }
    for(auto &q: qy[u]){
        ans[q.first] = query(root[u], q.second);
    }
}

void clear(Trie* rt){
    if(rt->next[0]) clear(rt->next[0]);
    if(rt->next[1]) clear(rt->next[1]);
    free(rt);
}

int main()
{
    while(scanf("%d%d", &n, &q) != EOF){
        init();
        for(int i = 1; i <= n; i++){
          scanf("%d", &arr[i]);
          qy[i].clear();
        }
        int v;
        for(int i = 2; i <= n; i++){
            scanf("%d", &v);
            add_edge(v, i);
        }
        int u, x;
        for(int i = 0; i < q; i++){
            scanf("%d%d", &u, &x);
            qy[u].push_back(make_pair(i, x));
        }
        dfs(1, -1);
        for(int i = 0; i < q; i++)
              printf("%d
", ans[i]);
        clear(root[1]);
    }

    return 0;
}