给定一个数组和一个数字 k，找到一个子序列，使得
- 子序列中元素的总和最大
- 子序列元素的索引至少相差 k
- 如果我们选择索引 i 处的元素使得 i + k + 1 >= N，那么我们不能选择任何其他元素作为子序列的一部分。因此我们需要决定是选择这个元素还是它之后的元素之一。
- 如果我们选择索引 i 处的元素使得 i + k + 1 < N，那么我们可以选择的下一个元素位于 i + k + 1 索引处。因此我们需要决定是选择这两个元素，还是移动到下一个相邻元素。
例子
Input : arr[] = {4, 5, 8, 7, 5, 4, 3, 4, 6, 5} k = 2 Output: 19 Explanation: The highest value is obtained if you pick indices 1, 4, 7, 10 giving 4 + 7 + 3 + 5 = 19 Input: arr[] = {50, 70, 40, 50, 90, 70, 60, 40, 70, 50} k = 2 Output: 230 Explanation: There are 10 elements and k = 2. If you select 2, 5, and 9 you get a total value of 230, which is the maximum possible.一个简单的解决方案是一个一个地考虑所有子序列。在每个子序列中，检查距离条件并返回最大和子序列。
一个有效的解决方案是使用动态规划。
有两种情况：
这两种情况可以写成：
Let MS[i] denotes the maximum sum of subsequence from i = N-2 to 0. Base Case: MS[N-1] = arr[N-1] If i + 1 + k >= N MS[i] = max(arr[i], MS[i+1]), Else MS[i] = max(arr[i] + MS[i+k+1], MS[i+1]) Evidently, the solution to the problem is to find MS[0].下面是实现：
C++
// CPP program to find maximum sum subsequence // such that elements are at least k distance // away. #includeusing namespace std; int maxSum(int arr[], int N, int k) { // MS[i] is going to store maximum sum // subsequence in subarray from arr[i] // to arr[n-1] int MS[N]; // We fill MS from right to left. MS[N - 1] = arr[N - 1]; for (int i = N - 2; i >= 0; i--) { if (i + k + 1 >= N) MS[i] = max(arr[i], MS[i + 1]); else MS[i] = max(arr[i] + MS[i + k + 1], MS[i + 1]); } return MS[0]; } // Driver code int main() { int N = 10, k = 2; int arr[] = { 50, 70, 40, 50, 90, 70, 60, 40, 70, 50 }; cout << maxSum(arr, N, k); return 0; }
Java
// Java program to find maximum sum subsequence // such that elements are at least k distance // away. import java.io.*; class GFG { static int maxSum(int arr[], int N, int k) { // MS[i] is going to store maximum sum // subsequence in subarray from arr[i] // to arr[n-1] int MS[] = new int[N]; // We fill MS from right to left. MS[N - 1] = arr[N - 1]; for (int i = N - 2; i >= 0; i--) { if (i + k + 1 >= N) MS[i] = Math.max(arr[i], MS[i + 1]); else MS[i] = Math.max(arr[i] + MS[i + k + 1], MS[i + 1]); } return MS[0]; } // Driver code public static void main(String[] args) { int N = 10, k = 2; int arr[] = { 50, 70, 40, 50, 90, 70, 60, 40, 70, 50 }; System.out.println(maxSum(arr, N, k)); } } // This code is contributed by Prerna Saini
Python3
# Python3 program to find maximum # sum subsequence such that elements # are at least k distance away. def maxSum(arr, N, k): # MS[i] is going to store maximum sum # subsequence in subarray from arr[i] # to arr[n-1] MS = [0 for i in range(N)] # We fill MS from right to left. MS[N - 1] = arr[N - 1] for i in range(N - 2, -1, -1): if (i + k + 1 >= N): MS[i] = max(arr[i], MS[i + 1]) else: MS[i] = max(arr[i] + MS[i + k + 1], MS[i + 1]) return MS[0] # Driver code N = 10; k = 2 arr = [ 50, 70, 40, 50, 90, 70, 60, 40, 70, 50 ] print(maxSum(arr, N, k)) # This code is contributed by Anant Agarwal.
C#
// C# program to find maximum sum // subsequence such that elements // are at least k distance away. using System; class GFG { static int maxSum(int []arr, int N, int k) { // MS[i] is going to store maximum sum // subsequence in subarray from arr[i] // to arr[n-1] int []MS = new int[N]; // We fill MS from right to left. MS[N - 1] = arr[N - 1]; for (int i = N - 2; i >= 0; i--) { if (i + k + 1 >= N) MS[i] = Math.Max(arr[i], MS[i + 1]); else MS[i] = Math.Max(arr[i] + MS[i + k + 1], MS[i + 1]); } return MS[0]; } // Driver code public static void Main() { int N = 10, k = 2; int []arr = { 50, 70, 40, 50, 90, 70, 60, 40, 70, 50 }; Console.WriteLine(maxSum(arr, N, k)); } } // This code is contributed by Anant Agarwal.
PHP
= 0; $i--) { if ($i + $k + 1 >= $N) $MS[$i] = max($arr[$i], $MS[$i + 1]); else $MS[$i] = max($arr[$i] + $MS[$i + $k + 1], $MS[$i + 1]); } return $MS[0]; } // Driver code $N = 10; $k = 2; $arr = array(50, 70, 40, 50, 90, 70, 60, 40, 70, 50); echo(maxSum($arr, $N, $k)); // This code is contributed by Ajit. ?>
输出：
230时间复杂度：O(n)
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