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 ;

}

巴特西

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

Query on A Tree

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