使用Mo算法的子数组中的不同元素

给定大小为n且查询数量为q的数组’a []’。每个查询可以用两个整数l和r表示。您的任务是在子数组l至r中打印不同整数的数量。
给定a [i] <= $10^6$
例子 ：

Input : a[] = {1, 1, 2, 1, 2, 3}
        q = 3
        0 4
        1 3
        2 5
Output : 2
         2
         3
In query 1, number of distinct integers
in a[0...4] is 2 (1, 2)
In query 2, number of distinct integers 
in a[1..3] is 2 (1, 2)
In query 3, number of distinct integers 
in a[2..5] is 3 (1, 2, 3)

Input : a[] = {7, 3, 5, 9, 7, 6, 4, 3, 2}
        q = 4
        1 5
        0 4
        0 7
        1 8
output : 5
         4
         6
         7

令a [0…n-1]为输入数组，q [0..m-1]为查询数组。
方法 ：

1. 以L值从0到0的查询的方式对所有查询进行排序$\sqrt(n) - 1$放在一起，然后来自$\sqrt(n)$至$2*\sqrt(n) - 1$ ， 等等。块中的所有查询均按R值的升序排序。
2. 初始化数组freq []的大小$10^6$与0。 freq []数组将所有元素的频率计数保持在给定范围内。
3. 以一种查询方式一个接一个地处理所有查询，即每个查询都使用上一个查询中计算出的不同元素数量和频率数组，并将结果存储在结构中。
   • 令“ curr_Diff_element”为先前查询的不同元素的数量。
   • 删除上一个查询的其他元素。例如，如果上一个查询为[0，8]而当前查询为[3，9]，则删除a [0]，a [1]和a [2]
   • 添加当前查询的新元素。在与上述相同的示例中，添加a [9]。
4. 按照与先前提供的顺序相同的顺序对查询进行排序，并打印其存储的结果

添加elements()

• 将要添加的元素(freq [a [i]])的频率增加1。
• 如果元素a [i]的频率为1，则将curr_diff_element加1，因为范围中已添加了1个新元素。

删除elements()

• 将要删除的元素(a [i])的频率降低1。
• 如果元素a [i]的频率为0。只需将curr_diff_element减少1，因为已从范围中完全删除1个元素。

注意：在此算法中，在步骤2中，R的索引变量最多更改O(n * $\sqrt(n)$ )次，而L的最大值最多改变O(m * $\sqrt(n)$ )次。所有这些限制都是可能的，因为排序的查询首先在$\sqrt(n)$尺寸。

预处理部分需要O(m Log m)时间。

处理所有查询需要O(n * $\sqrt(n)$ )+ O(米* $\sqrt(n)$ )= O((m + n)* $\sqrt(n)$ ) 时间。
下面是上述方法的实现：

// Program to compute no. of different elements
// of ranges for different range queries
#include 
using namespace std;
  
// Used in frequency array (maximum value of an
// array element).
const int MAX = 1000000;
  
// Variable to represent block size. This is made
// global so compare() of sort can use it.
int block;
  
// Structure to represent a query range and to store
// index and result of a particular query range
struct Query {
    int L, R, index, result;
};
  
// Function used to sort all queries so that all queries
// of same block are arranged together and within a block,
// queries are sorted in increasing order of R values.
bool compare(Query x, Query y)
{
    // Different blocks, sort by block.
    if (x.L / block != y.L / block)
        return x.L / block < y.L / block;
  
    // Same block, sort by R value
    return x.R < y.R;
}
  
// Function used to sort all queries in order of their
// index value so that results of queries can be printed
// in same order as of input
bool compare1(Query x, Query y)
{
    return x.index < y.index;
}
  
// calculate distinct elements of all query ranges.
// m is number of queries n is size of array a[].
void queryResults(int a[], int n, Query q[], int m)
{
    // Find block size
    block = (int)sqrt(n);
  
    // Sort all queries so that queries of same
    // blocks are arranged together.
    sort(q, q + m, compare);
  
    // Initialize current L, current R and current
    // different elements
    int currL = 0, currR = 0;
    int curr_Diff_elements = 0;
  
    // Initialize frequency array with 0
    int freq[MAX] = { 0 };
  
    // Traverse through all queries
    for (int i = 0; i < m; i++) {
          
        // L and R values of current range
        int L = q[i].L, R = q[i].R;
  
        // Remove extra elements of previous range.
        // For example if previous range is [0, 3]
        // and current range is [2, 5], then a[0] 
        // and a[1] are subtracted
        while (currL < L) {
              
            // element a[currL] is removed
            freq[a[currL]]--;
            if (freq[a[currL]] == 0) 
                curr_Diff_elements--;
              
            currL++;
        }
  
        // Add Elements of current Range
        // Note:- during addition of the left
        // side elements we have to add currL-1
        // because currL is already in range
        while (currL > L) {
            freq[a[currL - 1]]++;
  
            // include a element if it occurs first time
            if (freq[a[currL - 1]] == 1) 
                curr_Diff_elements++;
              
            currL--;
        }
        while (currR <= R) {
            freq[a[currR]]++;
  
            // include a element if it occurs first time
            if (freq[a[currR]] == 1) 
                curr_Diff_elements++;
              
            currR++;
        }
  
        // Remove elements of previous range. For example
        // when previous range is [0, 10] and current range
        // is [3, 8], then a[9] and a[10] are subtracted
        // Note:- Basically for a previous query L to R
        // currL is L and currR is R+1. So during removal
        // of currR remove currR-1 because currR was
        // never included
        while (currR > R + 1) {
  
            // element a[currL] is removed
            freq[a[currR - 1]]--;
  
            // if occurrence of a number is reduced
            // to zero remove it from list of 
            // different elements
            if (freq[a[currR - 1]] == 0) 
                curr_Diff_elements--;
              
            currR--;
        }
        q[i].result = curr_Diff_elements;
    }
}
  
// print the result of all range queries in
// initial order of queries
void printResults(Query q[], int m)
{
    sort(q, q + m, compare1);
    for (int i = 0; i < m; i++) {
        cout << "Number of different elements" << 
               " in range " << q[i].L << " to " 
             << q[i].R << " are " << q[i].result << endl;
    }
}
  
// Driver program
int main()
{
    int a[] = { 1, 1, 2, 1, 3, 4, 5, 2, 8 };
    int n = sizeof(a) / sizeof(a[0]);
    Query q[] = { { 0, 4, 0, 0 }, { 1, 3, 1, 0 },
                  { 2, 4, 2, 0 } };
    int m = sizeof(q) / sizeof(q[0]);
    queryResults(a, n, q, m);
    printResults(q, m);
    return 0;
}

输出：

Number of different elements in range 0 to 4 are 3
Number of different elements in range 1 to 3 are 2
Number of different elements in range 2 to 4 are 3

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