使用有序集和GNU C++ PBDS计算反转

给定N个整数的数组arr [] 。任务是找到反转数。如果arr [i]> arr [j]且i 两个元素arr [i]和arr [j]构成一个反转。

例子

Input: arr[] = {8, 4, 2, 1}
Output: 6
Explanation:
Given array has six inversions:

• (8, 4): arr[0] > arr[1] and 0 < 1
• (8, 2): arr[0] > arr[2] and 0 < 2
• (8, 1): arr[0] > arr[3] and 0 < 3
• (4, 2): arr[1] > arr[2] and 1 < 2
• (4, 1): arr[1] > arr[3] and 1 < 3
• (2, 1): arr[2] > arr[3] and 2 < 3

Input: arr[] = {2, 3}
Output: 0
Explanation:
There is no such pair exists such that arr[i] > arr[j] and i < j.

我们已经讨论了以下方法：

• 计算数组中的反转设置1(使用合并排序)
• 计算数组中的反转|设置2(使用自平衡BST)
• 使用C++ STL中的Set计数反转
• 计算数组中的反转|第3组(使用BIT)

在本文中，我们将讨论一种使用有序集和GNU C++ PBDS的方法。

方法：
我们将使用函数order_of_key(K) ，该函数在对数N时间内返回严格小于K的元素数。

1. 将数组的第一个元素插入Ordered_Set中。
2. 对于arr []中的所有其余元素，请执行以下操作：
   • 将当前元素插入Ordered_Set中。
   • 使用函数order_of_key(arr [i] +1)在Ordered_Set中找到严格小于当前元素+ 1的元素数。
   • Ordered_Set和order_of_key(current_element + 1)大小之间的差异将给出当前元素的反转计数。

   例如：

   arr[] = {8, 4, 2, 1}
   Ordered_Set S = {8}
   For remaining element in arr[]:
   At index 1, the element is  4
   S = {4, 8}
   key = order_of_key(5) = 1
   The difference between size of S and key gives the total 
   number of inversion count for that current element.
   inversion_count = S.size() - key =  2 - 1 = 1
   Inversion Pairs are: (8, 4)
   
   At index 2, the element is  2
   S = {2, 4, 8}
   key = order_of_key(3) = 1
   inversion_count = S.size() - key =  3 - 1 = 2
   Inversion Pairs are: (8, 2) and (4, 2)
   
   At index 3, the element is 1
   S = {1, 2, 4, 8}
   key = order_of_key(2) = 1
   inversion_count = S.size() - key =  4 - 1 = 3
   Inversion Pairs are: (8, 1), (4, 1) and (2, 1)
   
   Total inversion count = 1 + 2 + 3 = 6
   

下面是上述方法的实现：

// Ordered set in GNU C++ based
// approach for inversion count
#include 
#include 
#include 
using namespace __gnu_pbds;
using namespace std;
  
// Ordered Set Tree
typedef tree,
             rb_tree_tag,
             tree_order_statistics_node_update>
    ordered_set;
  
// Returns inversion count in
// arr[0..n-1]
int getInvCount(int arr[], int n)
{
    int key;
    // Intialise the ordered_set
    ordered_set set1;
  
    // Insert the first
    // element in set
    set1.insert(arr[0]);
  
    // Intialise inversion
    // count to zero
    int invcount = 0;
  
    // Finding the inversion
    // count for current element
    for (int i = 1; i < n; i++) {
        set1.insert(arr[i]);
  
        // Number of elements strictly
        // less than arr[i]+1
        key = set1.order_of_key(arr[i] + 1);
  
        // Difference between set size
        // and key will give the
        // inversion count
        invcount += set1.size() - key;
    }
    return invcount;
}
  
// Driver's Code
int main()
{
    int arr[] = { 8, 4, 2, 1 };
    int n = sizeof(arr) / sizeof(int);
  
    // Function call to count
    // inversion
    cout << getInvCount(arr, n);
    return 0;
}

输出：

6

时间复杂度： O(Nlog N)