长度最多为 K 的非重叠子数组的最大和

给定一个长度为 N 的整数数组 ‘arr’ 和一个整数 ‘k’，选择一些不重叠的子数组，使得每个子数组的长度至多为‘k’，没有两个子数组相邻并且所有子数组的总和所选子阵列的元素最大。
例子：

Input : arr[] = {-1, 2, -3, 4, 5}, k = 2
Output : 11
Sub-arrays that maximizes sum will be {{2}, {4, 5}}.
Thus, the answer will be 2+4+5 = 11.

Input :arr[] = {1, 1, 1, 1, 1}, k = 1
Output : 3

朴素的方法：一种简单的方法是递归地生成满足上述条件的元素的所有可能子集，并找到总和最大的子集。
时间复杂度：O(2 ^N )
高效的方法：更好的方法是使用动态规划。
假设我们处于索引“i”处。
让dp[i]定义为满足上述条件的子数组{i,n-1}的所有可能子集的元素的最大和。
我们将有“K+1”可能的选择，即

1. 拒绝“i”并求解“i+1”。
2. 选择子数组 {i} 并求解 ‘i+2’
3. 选择子数组 {i, i+1} 并求解 ‘i+3’

因此，递推关系将是：

dp[i] = max(dp[i+1], arr[i]+dp[i+2], arr[i]+arr[i+1]+dp[i+3],
        ...arr[i]+arr[i+1]+arr[i+2]...+arr[i+k-1] + dp[i+k+1])

下面是上述方法的实现：

C++

// C++ program to implement above approach
#include 
#define maxLen 10
using namespace std;
 
// Variable to store states of dp
int dp[maxLen];
 
// Variable to check if a given state has been solved
bool visit[maxLen];
 
// Function to find the maximum sum subsequence
// such that no two elements are adjacent
int maxSum(int arr[], int i, int n, int k)
{
    // Base case
    if (i >= n)
        return 0;
 
    // To check if a state has been solved
    if (visit[i])
        return dp[i];
    visit[i] = 1;
 
    // Variable to store
    // prefix sum for sub-array
    // {i, j}
    int tot = 0;
    dp[i] = maxSum(arr, i + 1, n, k);
 
    // Required recurrence relation
    for (int j = i; j < i + k and j < n; j++) {
        tot += arr[j];
        dp[i] = max(dp[i], tot +
                     maxSum(arr, j + 2, n, k));
    }
 
    // Returning the value
    return dp[i];
}
 
// Driver code
int main()
{
    // Input array
    int arr[] = { -1, 2, -3, 4, 5 };
 
    int k = 2;
 
    int n = sizeof(arr) / sizeof(int);
 
    cout << maxSum(arr, 0, n, k);
 
    return 0;
}

Java

// Java program to implement above approach
import java.io.*;
 
class GFG
{
     
    static int maxLen = 10;
     
    // Variable to store states of dp
    static int dp[] = new int[maxLen];
     
    // Variable to check
    // if a given state has been solved
    static boolean []visit = new boolean[maxLen];
     
    // Function to find the maximum sum subsequence
    // such that no two elements are adjacent
    static int maxSum(int arr[], int i,
                    int n, int k)
    {
        // Base case
        if (i >= n)
            return 0;
     
        // To check if a state has been solved
        if (visit[i])
            return dp[i];
        visit[i] = true;
     
        // Variable to store
        // prefix sum for sub-array
        // {i, j}
        int tot = 0;
        dp[i] = maxSum(arr, i + 1, n, k);
     
        // Required recurrence relation
        for (int j = i; j < (i + k) &&
                            (j < n); j++)
        {
            tot += arr[j];
            dp[i] = Math.max(dp[i], tot +
                    maxSum(arr, j + 2, n, k));
        }
     
        // Returning the value
        return dp[i];
    }
 
    // Driver code
    public static void main (String[] args)
    {
 
        // Input array
        int arr[] = { -1, 2, -3, 4, 5 };
         
        int k = 2;
         
        int n = arr.length;
         
        System.out.println(maxSum(arr, 0, n, k));
    }
}
 
// This code is contributed by ajit.

Python3

# Python3 program to implement above approach
maxLen = 10
 
# Variable to store states of dp
dp = [0]*maxLen;
 
# Variable to check if a given state has been solved
visit = [0]*maxLen;
 
# Function to find the maximum sum subsequence
# such that no two elements are adjacent
def maxSum(arr, i, n, k) :
 
    # Base case
    if (i >= n) :
        return 0;
 
    # To check if a state has been solved
    if (visit[i]) :
        return dp[i];
         
    visit[i] = 1;
 
    # Variable to store
    # prefix sum for sub-array
    # {i, j}
    tot = 0;
    dp[i] = maxSum(arr, i + 1, n, k);
 
    # Required recurrence relation
    j = i
    while (j < i + k and j < n) :
        tot += arr[j];
        dp[i] = max(dp[i], tot +
                    maxSum(arr, j + 2, n, k));
                     
        j += 1
     
    # Returning the value
    return dp[i];
 
 
# Driver code
if __name__ == "__main__" :
 
    # Input array
    arr = [ -1, 2, -3, 4, 5 ];
 
    k = 2;
 
    n = len(arr);
 
    print(maxSum(arr, 0, n, k));
     
# This code is contributed by AnkitRai01

C#

// C# program to implement above approach
using System;
 
class GFG
{
static int maxLen = 10;
 
// Variable to store states of dp
static int []dp = new int[maxLen];
 
// Variable to check
// if a given state has been solved
static bool []visit = new bool[maxLen];
 
// Function to find the maximum sum subsequence
// such that no two elements are adjacent
static int maxSum(int []arr, int i,
                  int n, int k)
{
    // Base case
    if (i >= n)
        return 0;
 
    // To check if a state has been solved
    if (visit[i])
        return dp[i];
    visit[i] = true;
 
    // Variable to store
    // prefix sum for sub-array
    // {i, j}
    int tot = 0;
    dp[i] = maxSum(arr, i + 1, n, k);
 
    // Required recurrence relation
    for (int j = i; j < (i + k) &&
                        (j < n); j++)
    {
        tot += arr[j];
        dp[i] = Math.Max(dp[i], tot +
                  maxSum(arr, j + 2, n, k));
    }
 
    // Returning the value
    return dp[i];
}
 
// Driver code
static public void Main ()
{
 
    // Input array
    int []arr = { -1, 2, -3, 4, 5 };
     
    int k = 2;
     
    int n = arr.Length;
     
    Console.WriteLine (maxSum(arr, 0, n, k));
}
}
 
// This code is contributed by ajit.

Javascript

    // Javascript program to implement above approach
     
    let maxLen = 10;
   
    // Variable to store states of dp
    let dp = new Array(maxLen);
 
    // Variable to check
    // if a given state has been solved
    let visit = new Array(maxLen);
 
    // Function to find the maximum sum subsequence
    // such that no two elements are adjacent
    function maxSum(arr, i, n, k)
    {
        // Base case
        if (i >= n)
            return 0;
 
        // To check if a state has been solved
        if (visit[i])
            return dp[i];
        visit[i] = true;
 
        // Variable to store
        // prefix sum for sub-array
        // {i, j}
        let tot = 0;
        dp[i] = maxSum(arr, i + 1, n, k);
 
        // Required recurrence relation
        for (let j = i; j < (i + k) &&
                            (j < n); j++)
        {
            tot += arr[j];
            dp[i] = Math.max(dp[i], tot +
                      maxSum(arr, j + 2, n, k));
        }
 
        // Returning the value
        return dp[i];
    }
     
    // Input array
    let arr = [ -1, 2, -3, 4, 5 ];
       
    let k = 2;
       
    let n = arr.length;
       
    document.write(maxSum(arr, 0, n, k));
 

输出：

11

时间复杂度： O(n*k)

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