范围查询以计算位于给定范围内的元素：MO的算法

给定一个由N个元素和两个整数A到B组成的数组arr [] ，任务是回答Q个查询，每个查询具有两个整数L和R。对于每个查询，找到子数组arr [L…R]中的元素数量，其中位于A到B (含)范围内。

例子：

Input: arr[] = {7, 3, 9, 13, 5, 4}, A = 4, B = 7
query = {1, 5}
Output: 2
Explanation:
Only 5 and 4 lies within 4 to 7
in the subarray {3, 9, 13, 5, 4}
Therefore, the count of such elements is 2.

Input: arr[] = {0, 1, 2, 3, 4, 5, 6, 7}, A = 1, B = 5
query = {3, 5}
Output: 3
Explanation:
All the elements 3, 4 and 5 lies within
the range 1 to 5 in the subarray {3, 4, 5}.
Therefore, the count of such elements is 3.

先决条件： MO的算法，SQRT分解

方法：想法是使用MO的算法对所有查询进行预处理，以便一个查询的结果可以在下一个查询中使用。下面是步骤说明：

• 将查询分组为多个块，其中每个块包含(0到√N– 1)，(√N到2x√N– 1)的起始范围值，依此类推。以R的升序对块内的查询进行排序。
• 以每个查询都使用上一个查询中计算出的结果的方式，一个接一个地处理所有查询。
• 保持频率数组，该数组将对出现在[L，R]范围内的arr [i]的频率进行计数。

  For example: arr[] = [3, 4, 6, 2, 7, 1], L = 0, R = 4 and A = 1, B = 6
  Initially frequency array is initialized to 0 i.e freq[]=[0….0]
  Step 1: Add arr[0] and increment its frequency as freq[arr[0]]++ i.e freq[3]++
  and freq[]=[0, 0, 0, 1, 0, 0, 0, 0]

  Step 2: Add arr[1] and increment freq[arr[1]]++ i.e freq[4]++
  and freq[]=[0, 0, 0, 1, 1, 0, 0, 0]

  Step 3: Add arr[2] and increment freq[arr[2]]++ i.e freq[6]++
  and freq[]=[0, 0, 0, 1, 1, 0, 1, 0]

  Step 4: Add arr[3] and increment freq[arr[3]]++ i.e freq[2]++
  and freq[]=[0, 0, 1, 1, 1, 0, 1, 0]

  Step 5: Add arr[4] and increment freq[arr[4]]++ i.e freq[7]++
  and freq[]=[0, 0, 1, 1, 1, 0, 1, 1]

  Step 6: Now we need to find the numbers of elements between A and B.

  Step 7: The answer is equal to $\sum_{i=A}^B freq[i]$

  To calculate the sum in step 7, we cannot do iteration because that would lead to O(N) time complexity per query so we will use sqrt decomposition technique to find the sum whose time complexity is O(√N) per query.

  下面是上述方法的实现：

  C++

  // C++ implementation to find the 
  // values in the range A to B 
  // in a subarray of L to R
    
  #include 
    
  using namespace std;
    
  #define MAX 100001
  #define SQRSIZE 400
    
  // Variable to represent block size.
  // This is made global so compare()
  // of sort can use it.
  int query_blk_sz;
    
  // Structure to represent a
  // query range
  struct Query {
      int L;
      int R;
  };
    
  // Frequency array
  // to keep count of elements
  int frequency[MAX];
    
  // Array which contains the frequency
  // of a particular block
  int blocks[SQRSIZE];
    
  // Block size
  int blk_sz;
    
  // Function used to sort all queries
  // so that all queries of the same
  // block are arranged together and
  // within a block, queries are sorted
  // in increasing order of R values.
  bool compare(Query x, Query y)
  {
      if (x.L / query_blk_sz != y.L / query_blk_sz)
          return x.L / query_blk_sz < y.L / query_blk_sz;
    
      return x.R < y.R;
  }
    
  // Function used to get the block
  // number of current a[i] i.e ind
  int getblocknumber(int ind)
  {
      return (ind) / blk_sz;
  }
    
  // Function to get the answer
  // of range [0, k] which uses the
  // sqrt decompostion technique
  int getans(int A, int B)
  {
      int ans = 0;
      int left_blk, right_blk;
      left_blk = getblocknumber(A);
      right_blk = getblocknumber(B);
    
      // If left block is equal to
      // rigth block then we can traverse
      // that block
      if (left_blk == right_blk) {
          for (int i = A; i <= B; i++)
              ans += frequency[i];
      }
      else {
          // Traversing first block in
          // range
          for (int i = A; i < (left_blk + 1) * blk_sz; i++)
              ans += frequency[i];
    
          // Traversing completely overlapped
          // blocks in range
          for (int i = left_blk + 1;
              i < right_blk; i++)
              ans += blocks[i];
    
          // Traversing last block in range
          for (int i = right_blk * blk_sz;
              i <= B; i++)
              ans += frequency[i];
      }
      return ans;
  }
    
  void add(int ind, int a[])
  {
      // Increment the frequency of a[ind]
      // in the frequency array
      frequency[a[ind]]++;
    
      // Get the block number of a[ind]
      // to update the result in blocks
      int block_num = getblocknumber(a[ind]);
    
      blocks[block_num]++;
  }
  void remove(int ind, int a[])
  {
      // Decrement the frequency of
      // a[ind] in the frequency array
      frequency[a[ind]]--;
    
      // Get the block number of a[ind]
      // to update the result in blocks
      int block_num = getblocknumber(a[ind]);
    
      blocks[block_num]--;
  }
  void queryResults(int a[], int n,
                  Query q[], int m, int A, int B)
  {
    
      // Initialize the block size
      // for queries
      query_blk_sz = sqrt(m);
    
      // Sort all queries so that queries
      // of same blocks are arranged
      // together.
      sort(q, q + m, compare);
    
      // Initialize current L,
      // current R and current result
      int currL = 0, currR = 0;
    
      for (int i = 0; i < m; i++) {
    
          // L and R values of the
          // current range
    
          int L = q[i].L, R = q[i].R;
    
          // Add Elements of current
          // range
          while (currR <= R) {
              add(currR, a);
              currR++;
          }
          while (currL > L) {
              add(currL - 1, a);
              currL--;
          }
    
          // Remove element of previous
          // range
          while (currR > R + 1)
    
          {
              remove(currR - 1, a);
              currR--;
          }
          while (currL < L) {
              remove(currL, a);
              currL++;
          }
          printf("%d\n", getans(A, B));
      }
  }
    
  // Driver code
  int main()
  {
    
      int arr[] = { 2, 0, 3, 1, 4, 2, 5, 11 };
      int N = sizeof(arr) / sizeof(arr[0]);
    
      int A = 1, B = 5;
      blk_sz = sqrt(N);
      Query Q[] = { { 0, 2 }, { 0, 3 }, { 5, 7 } };
    
      int M = sizeof(Q) / sizeof(Q[0]);
    
      // Answer the queries
      queryResults(arr, N, Q, M, A, B);
      return 0;
  }

  输出：

  2
  3
  2