HDU 6191 Query on A Tree(可持久化Trie+DFS序)

Query on A Tree

Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 132768/132768 K (Java/Others)
Total Submission(s): 712    Accepted Submission(s): 266

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?


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,Fn1Fi 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.



1Fin, the root of the tree is node 1.



For each query, just print an integer in a line indicating the largest result.


Sample Input
2 2
1 2
1 3
2 1


Sample Output

题目链接:HDU 6191

本来以为是很水的一道题目(确实很水),结果被坑在update的构建顺序上WA了很久,因为是可持久化,需要上一个版本的信息,而恰好一开始是先DFS序,再for每一个元素用它的L[i]时间戳来构建,其实这是有问题的, 因为上一个元素的L[i]不一定跟当前的L[i]连续,因此要按照L[i]的顺序来构建,而不是i的顺序;构建好之后查询一下L[u]~R[u]之间的与x的异或最大值即可


#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <cstdlib>
#include <cstring>
#include <bitset>
#include <string>
#include <stack>
#include <cmath>
#include <queue>
#include <set>
#include <climits>
#include <map>
using namespace std;
#define INF 0x3f3f3f3f
#define LC(x) (x<<1)
#define RC(x) ((x<<1)+1)
#define MID(x,y) ((x+y)>>1)
#define fin(name) freopen(name,"r",stdin)
#define fout(name) freopen(name,"w",stdout)
#define CLR(arr,val) memset(arr,val,sizeof(arr))
#define FAST_IO ios::sync_with_stdio(false);cin.tie(0);
typedef pair<int, int> pii;
typedef long long LL;
const double PI = acos(-1.0);
const int N = 100010;
struct edge
    int to, nxt;
    edge() {}
    edge(int _to, int _nxt): to(_to), nxt(_nxt) {}
} E[N];
struct Trie
    int nxt[2];
    int cnt;
    void init()
        nxt[0] = nxt[1] = 0;
        cnt = 0;
} L[N * 31];
int tot, root[N];
int head[N], etot;
int ll[N], rr[N], idx;
int arr[N];

void init()
    tot = 0;
    CLR(head, -1);
    etot = 0;
    idx = 0;
inline void add(int s, int t)
    E[etot] = edge(t, head[s]);
    head[s] = etot++;
void update(int &cur, int ori, int step, LL n, int v)
    cur = ++tot;
    L[cur] = L[ori];
    L[cur].cnt += v;
    if (step < 0)
        return ;
    int t = (n >> step) & 1;
    update(L[cur].nxt[t], L[ori].nxt[t], step - 1, n, v);
int Find(int S, int E, int step, LL n)
    if (step < 0)
        return 0;
    int t = (n >> step) & 1;
    if (L[L[E].nxt[t ^ 1]].cnt - L[L[S].nxt[t ^ 1]].cnt > 0)
        return (1LL << step) + Find(L[S].nxt[t ^ 1], L[E].nxt[t ^ 1], step - 1, n);
        return Find(L[S].nxt[t], L[E].nxt[t], step - 1, n);
void dfs_build(int u)
    ll[u] = ++idx;
    update(root[ll[u]], root[ll[u] - 1], 29, arr[u], 1);
    for (int i = head[u]; ~i; i = E[i].nxt)
        int v = E[i].to;
    rr[u] = idx;
int main(void)
    int n, q, i;
    while (~scanf("%d%d", &n, &q))
        for (i = 1; i <= n; ++i)
            scanf("%d", &arr[i]);
        for (i = 2; i <= n; ++i)
            int f;
            scanf("%d", &f);
            add(f, i);
        while (q--)
            int u, x;
            scanf("%d%d", &u, &x);
            int l = ll[u], r = rr[u];
            printf("%d\n", Find(root[l - 1], root[r], 29, x));
    return 0;
posted @ 2017-09-08 10:53  Blackops  阅读(...)  评论(...编辑  收藏