2017ACM/ICPC广西邀请赛 K- Query on A Tree trie树合并

Query on A Tree

Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 132768/132768 K (Java/Others)

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

题解：

　　每个数存在各自trie树里边，n个点这是棵树，再从底向上tri树合并起来

　　查询就是查询一颗合并后的trie树，利用从高位到低位，贪心取

#include <bits/stdc++.h>
inline int read(){int x=,f=;char ch=getchar();while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}while(ch>=''&&ch<=''){x=x*+ch-'';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*][], root[N],sz;
 
void inserts(int u,int x) {
    root[u] = ++sz;
    int tmp = sz;
    int y = sz;
    for(int i = ; i >= ; --i) {
        int tmps = (x>>i)&;
        if(!ch[y][tmps]) ch[y][tmps] = ++sz;
        y = ch[y][tmps];
    }
}
int merges(int u,int to) {
    if(u == ) return to;
    if(to == ) return u;
    int t = ++sz;
    ch[t][] = merges(ch[u][],ch[to][]);
    ch[t][] = merges(ch[u][],ch[to][]);
    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 = ;
    for(int i = ; i >= ; --i) {
        int tmps = (x>>i)&;
        if(ch[y][tmps^]) ret += (<<i),y = ch[y][tmps^];
        else y = ch[y][tmps];
    }
    return ret;
}
void init() {
    for(int i = ; i <= n; ++i) root[i] = ,G[i].clear();
    sz = ;
    memset(ch,,sizeof(ch));
}
int main( int argc , char * argv[] ){
    while(scanf("%d%d",&n,&q)!=EOF) {
        for(int i = ; i <= n; ++i) scanf("%d",&a[i]);
        init();
        for(int i = ; i <= n; ++i) {
            scanf("%d",&x);
            G[x].push_back(i);
        }
        dfs();
        for(int i = ; i <= q; ++i) {
            scanf("%d%d",&u,&x);
            printf("%lld\n",query(u,x));
        }
    }
    return ;
}