最大总和增加子序列| DP-14

给定n个正整数数组。编写程序以查找给定数组的最大和子序列的总和，以使子序列中的整数按升序排序。例如，如果输入为{1、101、2、3、100、4、5}，那么如果输入数组为{3、4、5、10，则输出应为106(1 + 2 + 3 + 100) }，则输出应为22(3 + 4 + 5 + 10)，如果输入数组为{10，5，4，3}，则输出应为10

解决方案
此问题是标准最长增加子序列(LIS)问题的变体。我们需要对LIS问题的动态编程解决方案稍作更改。我们需要更改的只是使用总和作为标准，而不是增加子序列的长度。

以下是该问题的动态编程解决方案：

C++

/* Dynamic Programming implementation 
of Maximum Sum Increasing Subsequence 
(MSIS) problem */
#include 
using namespace std;
  
/* maxSumIS() returns the maximum 
sum of increasing subsequence 
in arr[] of size n */
int maxSumIS(int arr[], int n) 
{ 
    int i, j, max = 0; 
    int msis[n]; 
  
    /* Initialize msis values 
    for all indexes */
    for ( i = 0; i < n; i++ ) 
        msis[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] && 
                msis[i] < msis[j] + arr[i]) 
                msis[i] = msis[j] + arr[i]; 
  
    /* Pick maximum of 
    all msis values */
    for ( i = 0; i < n; i++ ) 
        if ( max < msis[i] ) 
            max = msis[i]; 
  
    return max; 
} 
  
// Driver Code 
int main() 
{ 
    int arr[] = {1, 101, 2, 3, 100, 4, 5}; 
    int n = sizeof(arr)/sizeof(arr[0]); 
    cout << "Sum of maximum sum increasing "
            "subsequence is " << maxSumIS( arr, n ) << endl; 
    return 0; 
} 
  
// This is code is contributed by rathbhupendra

C

/* Dynamic Programming implementation 
of Maximum Sum Increasing Subsequence 
(MSIS) problem */
#include
  
/* maxSumIS() returns the maximum 
   sum of increasing subsequence
   in arr[] of size n */
int maxSumIS(int arr[], int n)
{
    int i, j, max = 0;
    int msis[n];
  
    /* Initialize msis values
       for all indexes */
    for ( i = 0; i < n; i++ )
        msis[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] && 
                msis[i] < msis[j] + arr[i])
                msis[i] = msis[j] + arr[i];
  
    /* Pick maximum of
       all msis values */
    for ( i = 0; i < n; i++ )
        if ( max < msis[i] )
            max = msis[i];
  
    return max;
}
  
// Driver Code
int main()
{
    int arr[] = {1, 101, 2, 3, 100, 4, 5};
    int n = sizeof(arr)/sizeof(arr[0]);
    printf("Sum of maximum sum increasing " 
            "subsequence is %d\n",
              maxSumIS( arr, n ) );
    return 0;
}

Java

/* Dynamic Programming Java 
   implementation of Maximum Sum
   Increasing Subsequence (MSIS)
   problem */
class GFG
{
    /* maxSumIS() returns the 
    maximum sum of increasing
    subsequence in arr[] of size n */
    static int maxSumIS(int arr[], int n)
    {
        int i, j, max = 0;
        int msis[] = new int[n];
  
        /* Initialize msis values 
           for all indexes */
        for (i = 0; i < n; i++)
            msis[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] &&
                    msis[i] < msis[j] + arr[i])
                    msis[i] = msis[j] + arr[i];
  
        /* Pick maximum of all
           msis values */
        for (i = 0; i < n; i++)
            if (max < msis[i])
                max = msis[i];
  
        return max;
    }
  
    // Driver code
    public static void main(String args[])
    {
        int arr[] = new int[]{1, 101, 2, 3, 100, 4, 5};
        int n = arr.length;
        System.out.println("Sum of maximum sum "+
                            "increasing subsequence is "+
                              maxSumIS(arr, n));
    }
}
  
// This code is contributed 
// by Rajat Mishra

Python

# Dynamic Programming bsed Python 
# implementation of Maximum Sum 
# Increasing Subsequence (MSIS)
# problem
  
# maxSumIS() returns the maximum 
# sum of increasing subsequence 
# in arr[] of size n
def maxSumIS(arr, n):
    max = 0
    msis = [0 for x in range(n)]
  
    # Initialize msis values
    # for all indexes
    for i in range(n):
        msis[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
                msis[i] < msis[j] + arr[i]):
                msis[i] = msis[j] + arr[i]
  
    # Pick maximum of
    # all msis values
    for i in range(n):
        if max < msis[i]:
            max = msis[i]
  
    return max
  
# Driver Code
arr = [1, 101, 2, 3, 100, 4, 5]
n = len(arr)
print("Sum of maximum sum increasing " + 
                     "subsequence is " +
                  str(maxSumIS(arr, n)))
  
# This code is contributed 
# by Bhavya Jain

C#

// Dynamic Programming C# implementation
// of Maximum Sum Increasing Subsequence
// (MSIS) problem 
using System;
class GFG {
      
    // maxSumIS() returns the 
    // maximum sum of increasing
    // subsequence in arr[] of size n 
    static int maxSumIS( int []arr, int n )
    {
        int i, j, max = 0;
        int []msis = new int[n];
  
        /* Initialize msis values
           for all indexes */
        for ( i = 0; i < n; i++ )
            msis[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] &&
                    msis[i] < msis[j] + arr[i])
                    msis[i] = msis[j] + arr[i];
  
        /* Pick maximum of all 
           msis values */
        for ( i = 0; i < n; i++ )
            if ( max < msis[i] )
                max = msis[i];
  
        return max;
    }
      
    // Driver Code
    public static void Main()
    {
        int []arr = new int[]{1, 101, 2, 3, 100, 4, 5};
        int n = arr.Length;
        Console.WriteLine("Sum of maximum sum increasing "+
                                        " subsequence is "+
        maxSumIS(arr, n));
    }
}
  
// This code is contributed by Sam007

PHP

 $arr[$j] && 
                $msis[$i] < $msis[$j] + $arr[$i])
                $msis[$i] = $msis[$j] + $arr[$i];
  
    // Pick maximum of all msis values
    for($i = 0;$i < $n; $i++ )
        if ($max < $msis[$i] )
            $max = $msis[$i];
  
    return $max;
}
  
    // Driver Code
    $arr = array(1, 101, 2, 3, 100, 4, 5);
    $n = count($arr);
    echo "Sum of maximum sum increasing subsequence is " 
                                   .maxSumIS( $arr, $n );
          
// This code is contributed by Sam007
?>

输出 ：

Sum of maximum sum increasing subsequence is 106

时间复杂度：O(n ^ 2)