子序列的最大和，以使两个元素在数组中的距离<K都不出现

给定n个整数和整数k的数组arr [] ，任务是找到子序列的最大和，以使子序列中的两个元素在原始数组中的距离≤k处均不出现。

例子：

Input: arr[] = {5, 3, 4, 11, 2}, k=1
Output: 16
All possible sub-sequences are {5, 4, 2}, {5, 11}, {5, 2}, {3, 11}, {3, 2}, {4, 2} and {11}
Out of which 5 + 11 = 16 gives the maximum sum.

Input: arr[] = {6, 7, 1, 3, 8, 2, 4}, k = 2
Output: 15

方法：在索引i处选择元素时，我们有两个选择，要么在子序列中包含当前元素，要么不包含。令dp [i]表示到目前为止到达索引i处的元素的最大和。我们可以如下计算dp [i]的值：

dp[i] = max(dp[i – (k + 1)] + arr[i], dp[i – 1])

dp[i – (k + 1)] + arr[i] is the case when element at index i is included. In that situation, maximum value will be arr[i] + maximum value till the last included element from the array.

dp[i – 1] is the case when current element is not included and the maximum value till now will be the maximum value till the previous element.

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
  
// Function to return the maximum sum possible
int maxSum(int* arr, int k, int n)
{
    if (n == 0)
        return 0;
    if (n == 1)
        return arr[0];
    if (n == 2)
        return max(arr[0], arr[1]);
  
    // dp[i] represent the maximum sum so far
    // after reaching current position i
    int dp[n];
  
    // Initialize dp[0]
    dp[0] = arr[0];
  
    // Initialize the dp values till k since any
    // two elements included in the sub-sequence
    // must be atleast k indices apart, and thus
    // first element and second element
    // will be k indices apart
    for (int i = 1; i <= k; i++)
        dp[i] = max(arr[i], dp[i - 1]);
  
    // Fill remaining positions
    for (int i = k + 1; i < n; i++)
        dp[i] = max(arr[i], dp[i - (k + 1)] + arr[i]);
  
    // Return the maximum sum
    int max = *(std::max_element(dp, dp + n));
    return max;
}
  
// Driver code
int main()
{
    int arr[] = { 6, 7, 1, 3, 8, 2, 4 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int k = 2;
    cout << maxSum(arr, k, n);
  
    return 0;
}

Java

// Java implementation of the approach
class GFG
{
      
// Function to return the maximum sum possible
static int maxSum(int []arr, int k, int n)
{
    if (n == 0)
        return 0;
    if (n == 1)
        return arr[0];
    if (n == 2)
        return Math.max(arr[0], arr[1]);
  
    // dp[i] represent the maximum sum so far
    // after reaching current position i
    int[] dp = new int[n];
  
    // Initialize dp[0]
    dp[0] = arr[0];
  
    // Initialize the dp values till k since any
    // two elements included in the sub-sequence
    // must be atleast k indices apart, and thus
    // first element and second element
    // will be k indices apart
    for (int i = 1; i <= k; i++)
        dp[i] = Math.max(arr[i], dp[i - 1]);
  
    // Fill remaining positions
    for (int i = k + 1; i < n; i++)
        dp[i] = Math.max(arr[i], dp[i - (k + 1)] + arr[i]);
  
    // Return the maximum sum
    return maximum(dp);
}
  
static int maximum(int[] arr)
{
    int max = Integer.MIN_VALUE;
    for(int i = 0; i < arr.length; i++) 
    {
        if(arr[i] > max) 
        {
            max = arr[i];
        }
    }
    return max;
}
  
// Driver code
public static void main (String[] args)
{
    int []arr = { 6, 7, 1, 3, 8, 2, 4 };
    int n = arr.length;
    int k = 2;
    System.out.println(maxSum(arr, k, n));
}
}
  
// This code is contributed by mits

Python3

# Python3 implementation of the approach 
  
# Function to return the 
# maximum sum possible 
def maxSum(arr, k, n) : 
      
    if (n == 0) :
        return 0; 
    if (n == 1) :
        return arr[0]; 
    if (n == 2) :
        return max(arr[0], arr[1]); 
  
    # dp[i] represent the maximum sum so far 
    # after reaching current position i 
    dp = [0] * n ; 
  
    # Initialize dp[0] 
    dp[0] = arr[0]; 
  
    # Initialize the dp values till k since any 
    # two elements included in the sub-sequence 
    # must be atleast k indices apart, and thus 
    # first element and second element 
    # will be k indices apart 
    for i in range(1, k + 1) : 
        dp[i] = max(arr[i], dp[i - 1]); 
  
    # Fill remaining positions 
    for i in range(k + 1, n) : 
        dp[i] = max(arr[i], 
                    dp[i - (k + 1)] + arr[i]); 
  
    # Return the maximum sum 
    max_element = max(dp); 
    return max_element; 
  
# Driver code 
if __name__ == "__main__" : 
    arr = [ 6, 7, 1, 3, 8, 2, 4 ]; 
    n = len(arr); 
    k = 2; 
      
    print(maxSum(arr, k, n)); 
      
# This code is contributed by Ryuga

C#

// C# implementation of the approach
using System;
using System.Linq;
  
class GFG
{
      
// Function to return the maximum sum possible
static int maxSum(int []arr, int k, int n)
{
    if (n == 0)
        return 0;
    if (n == 1)
        return arr[0];
    if (n == 2)
        return Math.Max(arr[0], arr[1]);
  
    // dp[i] represent the maximum sum so far
    // after reaching current position i
    int[] dp = new int[n];
  
    // Initialize dp[0]
    dp[0] = arr[0];
  
    // Initialize the dp values till k since any
    // two elements included in the sub-sequence
    // must be atleast k indices apart, and thus
    // first element and second element
    // will be k indices apart
    for (int i = 1; i <= k; i++)
        dp[i] = Math.Max(arr[i], dp[i - 1]);
  
    // Fill remaining positions
    for (int i = k + 1; i < n; i++)
        dp[i] = Math.Max(arr[i], dp[i - (k + 1)] + arr[i]);
  
    // Return the maximum sum
    int max = dp.Max();
    return max;
}
  
// Driver code
static void Main()
{
    int []arr = { 6, 7, 1, 3, 8, 2, 4 };
    int n = arr.Length;
    int k = 2;
    Console.WriteLine(maxSum(arr, k, n));
}
}
  
// This code is contributed by mits

输出：

15