查找可以通过从数组中删除元素获得的最大点数

给定具有N个元素和两个整数L和R的数组A ，其中， $1\leq a_{x} \leq 10^{5}$和$1\leq L \leq R \leq N$ 。您可以选择数组中的任何元素(假设为_x )并将其删除，还可以删除等于a _x +1 ， a _x +2 … a _x + R和a _x -1 ， a _x -2 …的所有元素。数组中的_x -L此步骤将花费_x点。任务是从阵列中删除所有元素后，使总成本最大化。

例子：

Input : 2 1 2 3 2 2 1
        L = 1, R = 1
Output : 8
We select 2 to delete, then (2-1)=1 and (2+1)=3 will need to be deleted, 
for given L and R range respectively.
Repeat this until 2 is completely removed. So, total cost = 2*4 = 8.

Input : 2 4 2 9 5
        L = 1, R = 2
Output : 18
We select 2 to delete, then 5 and then 9.
So total cost = 2*2 + 5 + 9 = 18.

方法：我们将找到所有元素的数量。现在假设选择了元素X，然后将删除[XL，X + R]范围内的所有元素。现在，我们从L和R中选择最小范围，并找到选择元素X时要删除的元素。当删除元素X时，我们的结果将是先前删除的元素的最大值。我们将使用动态编程来存储先前删除的元素的结果，并进一步使用它。

C++

// C++ program to find maximum cost after
// deleting all the elements form the array
#include 
using namespace std;
  
// function to return maximum cost obtained
int maxCost(int a[], int n, int l, int r)
{
  
    int mx = 0, k;
    // find maximum element of the array.
    for (int i = 0; i < n; ++i)
        mx = max(mx, a[i]);
  
    // initialize count of all elements to zero.
    int count[mx + 1];
    memset(count, 0, sizeof(count));
  
    // calculate frequency of all elements of array.
    for (int i = 0; i < n; i++)
        count[a[i]]++;
  
    // stores cost of deleted elements.
    int res[mx + 1];
    res[0] = 0;
  
    // selecting minimum range from L and R.
    l = min(l, r);
  
    for (int num = 1; num <= mx; num++) {
  
        // finds upto which elements are to be
        // deleted when element num is selected.
        k = max(num - l - 1, 0);
  
        // get maximum when selecting element num or not.
        res[num] = max(res[num - 1], num * count[num] + res[k]);
    }
  
    return res[mx];
}
  
// Driver program
int main()
{
    int a[] = { 2, 1, 2, 3, 2, 2, 1 }, l = 1, r = 1;
  
    // size of array
    int n = sizeof(a) / sizeof(a[0]);
  
    // function call to find maximum cost
    cout << maxCost(a, n, l, r);
  
    return 0;
}

Java

//Java program to find maximum cost after
//deleting all the elements form the array
  
public class GFG {
      
    //function to return maximum cost obtained
    static int maxCost(int a[], int n, int l, int r)
    {
  
     int mx = 0, k;
     // find maximum element of the array.
     for (int i = 0; i < n; ++i)
         mx = Math.max(mx, a[i]);
  
     // initialize count of all elements to zero.
     int[] count = new int[mx + 1];
     for(int i = 0; i < count.length; i++)
         count[i] = 0;
  
     // calculate frequency of all elements of array.
     for (int i = 0; i < n; i++)
         count[a[i]]++;
  
     // stores cost of deleted elements.
     int[] res = new int[mx + 1];
     res[0] = 0;
  
     // selecting minimum range from L and R.
     l = Math.min(l, r);
  
     for (int num = 1; num <= mx; num++) {
  
         // finds upto which elements are to be
         // deleted when element num is selected.
         k = Math.max(num - l - 1, 0);
  
         // get maximum when selecting element num or not.
         res[num] = Math.max(res[num - 1], num * count[num] + res[k]);
     }
  
     return res[mx];
    }
  
    //Driver program
    public static void main(String[] args) {
          
        int a[] = { 2, 1, 2, 3, 2, 2, 1 }, l = 1, r = 1;
  
         // size of array
         int n = a.length;
  
         // function call to find maximum cost
         System.out.println(maxCost(a, n, l, r));
    }
}

Python 3

# Python 3 Program to find maximum cost after 
# deleting all the elements form the array 
  
# function to return maximum cost obtained 
def maxCost(a, n, l, r) :
  
    mx = 0
  
    # find maximum element of the array.
    for i in range(n) :
        mx = max(mx, a[i])
  
    # create and initialize count of all elements to zero.
    count = [0] * (mx + 1)
  
    # calculate frequency of all elements of array.
    for i in range(n) :
        count[a[i]] += 1
  
    # stores cost of deleted elements.
    res = [0] * (mx + 1)
    res[0] = 0
  
    # selecting minimum range from L and R.
    l = min(l, r)
  
    for num in range(1, mx + 1) :
  
        # finds upto which elements are to be 
        # deleted when element num is selected.
        k = max(num - l - 1, 0)
  
        # get maximum when selecting element num or not. 
        res[num] = max(res[num - 1], num * count[num] + res[k])
  
    return res[mx]
  
# Driver code
if __name__ == "__main__" :
  
    a = [2, 1, 2, 3, 2, 2, 1 ]
    l, r = 1, 1
  
    # size of array 
    n =  len(a)
  
    # function call to find maximum cost 
    print(maxCost(a, n, l, r))
  
# This code is contributed by ANKITRAI1

C#

// C# program to find maximum cost 
// after deleting all the elements 
// form the array
using System;
  
class GFG 
{
  
// function to return maximum 
// cost obtained
static int maxCost(int []a, int n, 
                   int l, int r)
{
    int mx = 0, k;
      
    // find maximum element
    // of the array.
    for (int i = 0; i < n; ++i)
        mx = Math.Max(mx, a[i]);
      
    // initialize count of all
    // elements to zero.
    int[] count = new int[mx + 1];
    for(int i = 0; i < count.Length; i++)
        count[i] = 0;
      
    // calculate frequency of all
    // elements of array.
    for (int i = 0; i < n; i++)
        count[a[i]]++;
      
    // stores cost of deleted elements.
    int[] res = new int[mx + 1];
    res[0] = 0;
      
    // selecting minimum range
    // from L and R.
    l = Math.Min(l, r);
      
    for (int num = 1; num <= mx; num++) 
    {
      
        // finds upto which elements 
        // are to be deleted when 
        // element num is selected.
        k = Math.Max(num - l - 1, 0);
      
        // get maximum when selecting 
        // element num or not.
        res[num] = Math.Max(res[num - 1], num *
                          count[num] + res[k]);
    }
  
return res[mx];
}
  
// Driver Code
public static void Main()
{
    int []a = { 2, 1, 2, 3, 2, 2, 1 };
    int l = 1, r = 1;
  
    // size of array
    int n = a.Length;
  
    // function call to find maximum cost
    Console.WriteLine(maxCost(a, n, l, r));
}
}
  
// This code is contributed 
// by inder_verma

输出：

8

时间复杂度： O(max(A))