在多个查询的子数组中查找不同（唯一）值的数量

Question

我有一个数组（可以有 2X10^5 个值）。我想对该数组执行大量查询。每个查询都是 [L,R] 类型，这个查询的结果应该是子数组中从索引 L 开始到索引 R 结束的唯一值的数量。

我知道这可以使用 Mo 的算法在 O(nrootn) 时间内完成。然而问题在于 Mo 的算法是一种离线算法。我正在寻找的是一个在线算法，因为上一个查询的结果决定了我的下一个查询。

我试着用它来形成一个线段树，其中的节点将存储范围内的所有不同元素。然而，这对我的目的来说太慢了。这种方法的预处理花费了太多时间。

Answer 1

这是我的 C++ 尝试解决方案（也发布了 here) using a Wavelet tree, implemented with code adapted from https://www.geeksforgeeks.org/wavelet-trees-introduction. The idea to reformulate the problem (as Photon 一个 link 到）是首先构造一个数组，列出原始数组中每个对应的单元格，下一个单元格的索引右边的重复元素。那么问题就变成了找到区间中有多少单元格具有超出当前区间的"next-index"（那些显然在区间内没有重复），可以用装饰小波树查询。请参阅底部的（非零基）查询示例。

// Adapted from https://www.geeksforgeeks.org/wavelet-trees-introduction

#include <iostream>
#include <vector>
#include <map>
#include <algorithm>
#include <climits>
using namespace std;

// wavelet tree class 
class wavelet_tree { 
public: 
    // Range to elements 
    int low, high; 

    // Left and Right child 
    wavelet_tree* l, *r; 

    std::vector<int> freq;

    // Default constructor 
    // Array is in range [x, y] 
    // Indices are in range [from, to] 
    wavelet_tree(int* from, int* to, int x, int y) 
    { 
        // Initialising low and high 
        low = x, high = y; 

        // Array is of 0 length 
        if (from >= to) 
            return; 

        // Array is homogenous 
        // Example : 1 1 1 1 1 
        if (high == low) { 
            // Assigning storage to freq array 
            freq.reserve(to - from + 1); 

            // Initialising the Freq array 
            freq.push_back(0); 

            // Assigning values 
            for (auto it = from; it != to; it++) 

                // freq will be increasing as there'll 
                // be no further sub-tree 
                freq.push_back(freq.back() + 1); 

            return; 
        } 

        // Computing mid 
        int mid = (low + high) / 2; 

        // Lambda function to check if a number 
        // is less than or equal to mid 
        auto lessThanMid = [mid](int x) { 
            return x <= mid; 
        }; 

        // Assigning storage to freq array 
        freq.reserve(to - from + 1); 

        // Initialising the freq array 
        freq.push_back(0); 

        // Assigning value to freq array 
        for (auto it = from; it != to; it++) 

            // If lessThanMid returns 1(true), we add 
            // 1 to previous entry. Otherwise, we add 0 
            // (element goes to right sub-tree) 
            freq.push_back(freq.back() + lessThanMid(*it));      

        // std::stable_partition partitions the array w.r.t Mid 
        auto pivot = std::stable_partition(from, to, lessThanMid); 

        // Left sub-tree's object 
        l = new wavelet_tree(from, pivot, low, mid); 

        // Right sub-tree's object 
        r = new wavelet_tree(pivot, to, mid + 1, high); 
    } 

    // Count of numbers in range[L..R] less than 
    // or equal to k 
    int kOrLess(int l, int r, int k) 
    { 
        // No elements int range is less than k 
        if (l > r or k < low) 
            return 0; 

        // All elements in the range are less than k 
        if (high <= k) 
            return r - l + 1; 

        // Computing LtCount and RtCount 
        int LtCount = freq[l - 1]; 
        int RtCount = freq[r]; 

        // Answer is (no. of element <= k) in 
        // left + (those <= k) in right 
        return (this->l->kOrLess(LtCount + 1, RtCount, k) + 
            this->r->kOrLess(l - LtCount, r - RtCount, k)); 
    } 

    // Count of numbers in range[L..R] greater than 
    // or equal to k 
    int kOrMore(int l, int r, int k) 
    { 
        // No elements int range are greater than k 
        if (l > r or k > high) 
            return 0; 

        // All elements in the range are greater than k 
        if (low >= k) 
            return r - l + 1; 

        // Computing LtCount and RtCount 
        int LtCount = freq[l - 1]; 
        int RtCount = freq[r]; 

        // Answer is (no. of element <= k) in 
        // left + (those <= k) in right 
        return (this->l->kOrMore(LtCount + 1, RtCount, k) + 
            this->r->kOrMore(l - LtCount, r - RtCount, k)); 
    }

}; 


int main() 
{ 
    int size = 7, high = INT_MIN;
    int arr[] = {1, 2, 3, 2, 4, 3, 1};
    int next[size];
    std::map<int, int> next_idx;

    for (int i=size-1; i>=0; i--){
        if (next_idx.find(arr[i]) == next_idx.end())
            next[i] = size + 1;
        else
            next[i] = next_idx[arr[i]];
        next_idx[arr[i]] = i + 1;
        high = max(high, next[i]);
    } 

    // Object of class wavelet tree 
    wavelet_tree obj(next, next + size, 1, high);

    // Queries are NON-zero-based
    //
    //  1  2  3  4  5  6  7
    // {1, 2, 3, 2, 4, 3, 1};
    // query([3, 6]) = 3;
    cout << obj.kOrMore(3, 6, 7) << '\n';
    // query([1, 4]) = 3;
    cout << obj.kOrMore(1, 4, 5) << '\n';
    // query([1, 7]) = 4;
    cout << obj.kOrMore(1, 7, 8) << '\n';

    return 0; 
}

在多个查询的子数组中查找不同（唯一）值的数量

Finding number of distinct (unique) values in a sub-array for multiple queries

c++

arrays

algorithm

data-structures

wavelet