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=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 = 30; 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;
}