给定一个由 N 个正整数组成的数组。任务是找到给定数组的最大和递减子序列(MSDS)的总和，使得子序列中的整数按降序排序。
例子：
Input: arr[] = {5, 4, 100, 3, 2, 101, 1}
Output: 106
100 + 3 + 2 + 1 = 106
Input: arr[] = {10, 5, 4, 3}
Output: 22
10 + 5 + 4 + 3 = 22
此问题是最长递减子序列问题的变体。上述问题的最优子结构将是：
设 arr[0..n-1] 是输入数组， MSDS[i] 是在索引 i 处结束的 MSDS 的最大总和，这样 arr[i] 是 MSDS 的最后一个元素。
那么，MSDS[i] 可以写成：
MSDS[i] = a[i] + max( MSDS[j] ) where i > j > 0 and arr[j] > arr[i] or,
MSDS[i] = a[i], if no such j exists.
要找到给定数组的 MSDS，我们需要返回 max(MSDS[i]) 其中 n > i > 0。
下面是上述方法的实现：
C++
// CPP code to return the maximum sum
// of decreasing subsequence in arr[]
#include
using namespace std;
// function to return the maximum
// sum of decreasing subsequence
// in arr[]
int maxSumDS(int arr[], int n)
{
int i, j, max = 0;
int MSDS[n];
// Initialize msds values
// for all indexes
for (i = 0; i < n; i++)
MSDS[i] = arr[i];
// Compute maximum sum values
// in bottom up manner
for (i = 1; i < n; i++)
for (j = 0; j < i; j++)
if (arr[i] < arr[j] && MSDS[i] < MSDS[j] + arr[i])
MSDS[i] = MSDS[j] + arr[i];
// Pick maximum of all msds values
for (i = 0; i < n; i++)
if (max < MSDS[i])
max = MSDS[i];
return max;
}
// Driver Code
int main()
{
int arr[] = { 5, 4, 100, 3, 2, 101, 1 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << "Sum of maximum sum decreasing subsequence is: "
<< maxSumDS(arr, n);
return 0;
} Java
// Java code to return the maximum sum
// of decreasing subsequence in arr[]
import java.io.*;
import java.lang.*;
class GfG {
// function to return the maximum
// sum of decreasing subsequence
// in arr[]
public static int maxSumDS(int arr[], int n)
{
int i, j, max = 0;
int[] MSDS = new int[n];
// Initialize msds values
// for all indexes
for (i = 0; i < n; i++)
MSDS[i] = arr[i];
// Compute maximum sum values
// in bottom up manner
for (i = 1; i < n; i++)
for (j = 0; j < i; j++)
if (arr[i] < arr[j] &&
MSDS[i] < MSDS[j] + arr[i])
MSDS[i] = MSDS[j] + arr[i];
// Pick maximum of all msds values
for (i = 0; i < n; i++)
if (max < MSDS[i])
max = MSDS[i];
return max;
}
// Driver Code
public static void main(String argc[])
{
int arr[] = { 5, 4, 100, 3, 2, 101, 1 };
int n = 7;
System.out.println("Sum of maximum sum"
+ " decreasing subsequence is: "
+ maxSumDS(arr, n));
}
}
// This code os contributed by Sagar Shukla.Python3
# Python3 code to return the maximum sum
# of decreasing subsequence in arr[]
# Function to return the maximum
# sum of decreasing subsequence
# in arr[]
def maxSumDS(arr, n):
i, j, max = (0, 0, 0)
MSDS=[0 for i in range(n)]
# Initialize msds values
# for all indexes
for i in range(n):
MSDS[i] = arr[i]
# Compute maximum sum values
# in bottom up manner
for i in range(1, n):
for j in range(i):
if (arr[i] < arr[j] and
MSDS[i] < MSDS[j] + arr[i]):
MSDS[i] = MSDS[j] + arr[i]
# Pick maximum of all msds values
for i in range(n):
if (max < MSDS[i]):
max = MSDS[i]
return max
if __name__ == "__main__":
arr=[5, 4, 100, 3,
2, 101, 1]
n=len(arr)
print("Sum of maximum sum decreasing subsequence is: ",
maxSumDS(arr, n))
# This code is contributed by Rutvik_56C#
// C# code to return the
// maximum sum of decreasing
// subsequence in arr[]
using System;
class GFG
{
// function to return the
// maximum sum of decreasing
// subsequence in arr[]
public static int maxSumDS(int []arr,
int n)
{
int i, j, max = 0;
int[] MSDS = new int[n];
// Initialize msds values
// for all indexes
for (i = 0; i < n; i++)
MSDS[i] = arr[i];
// Compute maximum sum values
// in bottom up manner
for (i = 1; i < n; i++)
for (j = 0; j < i; j++)
if (arr[i] < arr[j] &&
MSDS[i] < MSDS[j] + arr[i])
MSDS[i] = MSDS[j] + arr[i];
// Pick maximum of
// all msds values
for (i = 0; i < n; i++)
if (max < MSDS[i])
max = MSDS[i];
return max;
}
// Driver Code
static public void Main ()
{
int []arr = {5, 4, 100,
3, 2, 101, 1};
int n = 7;
Console.WriteLine("Sum of maximum sum" +
" decreasing subsequence is: " +
maxSumDS(arr, n));
}
}
// This code is contributed by m_kitPHP
Javascript
输出：
Sum of maximum sum decreasing subsequence is: 106时间复杂度： O(N 2 )
辅助空间： O(N)
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