为K个查询重新排列数组后的最大和

给定两个阵列的常用3包含N个整数和Q [] []包含K的查询，其中每个查询表示范围[L，R] []。任务是重新排列数组，并找到所有子数组的最大可能和，其中每个子数组由数组元素在每个查询给定的[L，R]范围内定义。

注意：在Q [] []数组中使用基于1的索引来表示范围。

例子：

Input: arr[] = { 2, 6, 10, 1, 5, 6 }, Q[][2] = {{1, 3}, {4, 6}, {3, 4}}
Output: 46
Explanation:
One possible way is to rearrange the array to arr[] = {2, 6, 10, 6, 5, 1}.
In this arrangement:
The sum of the subarray in the range [1, 3] = 2 + 6 + 10 = 18.
The sum of the subarray in the range [4, 6] = 6 + 5 + 1 = 12.
The sum of the subarray in the range [3, 4] = 10 + 6 = 16.
The total sum of all the subarrays = 46 which is the maximum possible.

Input: arr[] = { 1, 2, 3, 4, 5, 6, 7, 8 }, Q[][2] = {{1, 4}, {5, 5}, {7, 8}, {8, 8}}
Output: 43
Explanation:
One possible way is to rearrange the array to arr[] = {2, 3, 4, 5, 6, 1, 7, 8}.
In this arrangement:
The sum of the subarray in the range [1, 4] = 2 + 3 + 4 + 5 = 14.
The sum of the subarray in the range [5, 5] = 6 = 6.
The sum of the subarray in the range [7, 8] = 7 + 8 = 15.
The sum of the subarray in the range [8, 8] = 8 = 8.
The total sum of all the subarrays = 43 which is the maximum possible.

方法：观察清楚，可以得出的结论是，当最大元素包含在尽可能多的子数组中时，我们得到最大和。为此，我们需要通过迭代所有查询来找到包含每个索引的次数。

例如：假设数组为arr [] = {2，6，10，6，5，1} ，查询为Q [] [] = {{1，3}，{4，6}，{3， 4}} 。

1. 步骤1：创建一个大小为N的计数数组C [] 。因此，最初，计数数组C [] = {0，0，0，0，0，0} 。
2. 步骤2：对于查询[1，3] ，索引[ 1，3]处的元素增加1。此查询后的计数数组变为{1，1，1，1，0，0，0} 。
3. 步骤3：类似地，对于下一个查询，count数组变为{1，1，1，1，1，1，1}，最后，在第三个查询之后，count数组变为{1，1，2，2，1， 1} 。
4. 步骤4：获得count数组后，其想法是使用排序以获得最大的总和。
5. 步骤5：排序后，数组C [] = { 1，1，1，1，1，2，2 }和arr [] = {1,2,5,6,6,10} 。最大可能的总和是两个数组的加权总和，即：

   sum = ((1 * 1) + (1 * 2) + (1 * 5) + (1 * 6) + (2 * 6) + (2 * 10)) = 46

下面是上述方法的实现：

C++

// C++ program to find the maximum sum
// after rearranging the array for K queries
  
#include 
using namespace std;
  
// Function to find maximum sum after
// rearranging array elements
int maxSumArrangement(int A[], int R[][2],
                      int N, int M)
{
  
    // Auxiliary array to find the
    // count of each selected elements
    int count[N];
  
    // Initialize with 0
    memset(count, 0, sizeof count);
  
    // Finding count of every element
    // to be selected
    for (int i = 0; i < M; ++i) {
  
        int l = R[i][0], r = R[i][1] + 1;
  
        // Making it to 0-indexing
        l--;
        r--;
  
        // Prefix sum array concept is used
        // to obtain the count array
        count[l]++;
  
        if (r < N)
            count[r]--;
    }
  
    // Iterating over the count array
    // to get the final array
    for (int i = 1; i < N; ++i) {
        count[i] += count[i - 1];
    }
  
    // Variable to store the maximum sum
    int ans = 0;
  
    // Sorting both the arrays
    sort(count, count + N);
    sort(A, A + N);
  
    // Loop to find the maximum sum
    for (int i = N - 1; i >= 0; --i) {
        ans += A[i] * count[i];
    }
  
    return ans;
}
  
// Driver code
int main()
{
    int A[] = { 2, 6, 10, 1, 5, 6 };
    int R[][2]
        = { { 1, 3 }, { 4, 6 }, { 3, 4 } };
  
    int N = sizeof(A) / sizeof(A[0]);
    int M = sizeof(R) / sizeof(R[0]);
  
    cout << maxSumArrangement(A, R, N, M);
  
    return 0;
}

Java

// Java program to find the maximum sum
// after rearranging the array for K queries
import java.util.*;
  
class GFG
{
       
    // Function to find maximum sum after
    // rearranging array elements
    static int maxSumArrangement(int A[], int R[][],
                          int N, int M)
    {
       
        // Auxiliary array to find the
        // count of each selected elements
        int count[] = new int[N];
        int i;
       
        // Finding count of every element
        // to be selected
        for ( i = 0; i < M; ++i) {
       
            int l = R[i][0], r = R[i][1] + 1;
       
            // Making it to 0-indexing
            l--;
            r--;
       
            // Prefix sum array concept is used
            // to obtain the count array
            count[l]++;
       
            if (r < N)
                count[r]--;
        }
       
        // Iterating over the count array
        // to get the final array
        for (i = 1; i < N; ++i) {
            count[i] += count[i - 1];
        }
       
        // Variable to store the maximum sum
        int ans = 0;
       
        // Sorting both the arrays
        Arrays.sort( count);
        Arrays.sort(A);
       
        // Loop to find the maximum sum
        for (i = N - 1; i >= 0; --i) {
            ans += A[i] * count[i];
        }
       
        return ans;
    }
       
    // Driver code
    public static void main(String []args)
    {
        int A[] = { 2, 6, 10, 1, 5, 6 };
        int R[][]
            = { { 1, 3 }, { 4, 6 }, { 3, 4 } };
       
        int N = A.length;
        int M = R.length;
       
        System.out.print(maxSumArrangement(A, R, N, M));
       
    }
}
  
// This code is contributed by chitranayal

Python3

# Python3 program to find the maximum sum 
# after rearranging the array for K queries 
  
# Function to find maximum sum after 
# rearranging array elements 
def maxSumArrangement( A,  R,  N,  M):
  
    # Auxiliary array to find the 
    # count of each selected elements 
    # Initialize with 0 
    count = [0 for i in range(N)]
  
    # Finding count of every element 
    # to be selected 
    for i in range(M):
  
        l = R[i][0]
        r = R[i][1] + 1
  
        # Making it to 0-indexing 
        l = l - 1 
        r = r - 1
  
        # Prefix sum array concept is used 
        # to obtain the count array 
        count[l] = count[l] + 1 
  
        if (r < N):
            count[r] = count[r] - 1 
  
    # Iterating over the count array 
    # to get the final array 
    for i in range(1, N): 
        count[i] = count[i] + count[i - 1]
  
    # Variable to store the maximum sum 
    ans = 0
  
    # Sorting both the arrays 
    count.sort()
    A.sort()
  
    # Loop to find the maximum sum 
    for i in range(N - 1, -1, -1): 
        ans = ans + A[i] * count[i]
  
    return ans
  
# Driver code 
A = [ 2, 6, 10, 1, 5, 6 ]
R = [ [ 1, 3 ], [ 4, 6 ], [ 3, 4 ] ] 
  
N = len(A)
M = len(R)
  
print(maxSumArrangement(A, R, N, M))
  
# This code is contributed by Sanjit_Prasad

C#

// C# program to find the maximum sum
// after rearranging the array for K queries
using System;
  
class GFG
{
        
    // Function to find maximum sum after
    // rearranging array elements
    static int maxSumArrangement(int []A, int [,]R,
                          int N, int M)
    {
        
        // Auxiliary array to find the
        // count of each selected elements
        int []count = new int[N];
        int i;
        
        // Finding count of every element
        // to be selected
        for ( i = 0; i < M; ++i) {
        
            int l = R[i, 0], r = R[i, 1] + 1;
        
            // Making it to 0-indexing
            l--;
            r--;
        
            // Prefix sum array concept is used
            // to obtain the count array
            count[l]++;
        
            if (r < N)
                count[r]--;
        }
        
        // Iterating over the count array
        // to get the readonly array
        for (i = 1; i < N; ++i) {
            count[i] += count[i - 1];
        }
        
        // Variable to store the maximum sum
        int ans = 0;
        
        // Sorting both the arrays
        Array.Sort( count);
        Array.Sort(A);
        
        // Loop to find the maximum sum
        for (i = N - 1; i >= 0; --i) {
            ans += A[i] * count[i];
        }
        
        return ans;
    }
        
    // Driver code
    public static void Main(String []args)
    {
        int []A = { 2, 6, 10, 1, 5, 6 };
        int [,]R
            = { { 1, 3 }, { 4, 6 }, { 3, 4 } };
        
        int N = A.Length;
        int M = R.GetLength(0);
        
        Console.Write(maxSumArrangement(A, R, N, M));      
    }
}
  
// This code is contributed by Princi Singh

输出：

46

时间复杂度： O(N * log(N))