HDU

Query on A Tree

 HDU - 6191

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

HDU - 6191

2017ACM/ICPC广西邀请赛-重现赛（感谢广西大学）

My Solution

题意：给出一颗树，每个节点有一个权值，q个询问，询问以点u为根的子树中的节点权值异或x所得的值的最大值。

可持久化字典树+dfs序

对子树进行询问很容易想到dfs序，然后变成了线性的询问区间[l，r]内异或x所得的最大值，这个是可以用可持久化字典树来做，类似于主席树，每颗字典树维护的是区间[1,i]的信息，询问的时候，如果sum[r]-sum[l-1]大于0则说明在这个区间内这个值是存在的，然后按照通常的0-1树的做法，每次尽可能访问x&(1<<i)的异或值即可是答案尽可能大。

空间复杂度 O(nlogn)

时间复杂度 O(nlogn)

#include <iostream>#include <cstdio>#include <cstring>#include <vector>using namespace std;typedef long long LL;const int MAXN = 1e5 + 8;//可持久化字典树int root[MAXN];struct Persistent_Trie{    int tot;    int ch[32*MAXN][2], sum[32*MAXN]; //这里应该开logn+2倍    void init(){        tot = 0;        memset(ch, 0, sizeof ch);        memset(sum, 0, sizeof sum);    }    void _insert(int &rt, int last, int val){        int state, i, t;        rt = state = ++tot;        for(i = 30; i >= 0; i--){            ch[state][0] = ch[last][0];            ch[state][1] = ch[last][1];            sum[state] = sum[last] + 1;            t = val & (1 << i); t >>= i;            last = ch[last][t];            ch[state][t] = ++tot;            state = ch[state][t];        }        sum[state] = sum[last] + 1;    }    int query(int ss, int tt, int val){        int res = 0, i;        for(i = 30; i >= 0; i--){            int t = val & (1 << i); t >>= i;            if(sum[ch[tt][t^1]] - sum[ch[ss][t^1]]){                res += (1 << i);                tt = ch[tt][t^1];                ss = ch[ss][t^1];            }            else tt = ch[tt][t], ss = ch[ss][t];        }        return res;    }} pt;int a[MAXN];//dfs序vector<int> sons[MAXN];int p1[MAXN], p2[MAXN], ti = 0;int dfsnum[MAXN];  //这个按情况是否需要。inline void get_dfs_list(int u, int fa){    p1[u] = ++ti;    dfsnum[ti] = u; //    int sz = sons[u].size(), i, v;    for(i = 0; i < sz; i++){        v = sons[u][i];        if(v == fa) continue;        get_dfs_list(v, u);    }    p2[u] = ti;}int main(){    //cout << (1<<30) << " " << (1 << 29) << endl;    #ifdef LOCAL    freopen("j.txt", "r", stdin);    //freopen("j.out", "w", stdout);    int T = 1;    while(T--){    #endif // LOCAL    //ios::sync_with_stdio(false); cin.tie(0);    int n, q, i, u, x;    while(scanf("%d%d", &n, &q) != EOF){        pt.init();        for(i = 1; i <= n; i++){            scanf("%d", &a[i]);        }        for(i = 1; i < n; i++){            scanf("%d", &u);            sons[u].push_back(i + 1);        }        ti = 0;        get_dfs_list(1, 0);        for(i = 1; i <= ti; i++){            pt._insert(root[i], root[i-1], a[dfsnum[i]]);        }        while(q--){            scanf("%d%d", &u, &x);            printf("%d\n", pt.query(root[p1[u]-1], root[p2[u]], x));        }        for(i = 1; i <= n; i++) sons[i].clear();    }    #ifdef LOCAL    cout << endl;    }    #endif // LOCAL    return 0;}

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