给定一个由 n 个正整数组成的数组。编写一个程序，求给定数组的最大和子序列之和，使子序列中的整数按升序排列。例如，如果输入是{1, 101, 2, 3, 100, 4, 5}，那么输出应该是106 (1 + 2 + 3 + 100)，如果输入数组是{3, 4, 5, 10 }，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
此问题是标准最长递增子序列 (LIS) 问题的变体。我们需要对 LIS 问题的动态规划解决方案稍作改动。我们需要改变的只是使用 sum 作为标准而不是增加子序列的长度。
以下是该问题的动态规划解决方案：
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 MishraPython
# 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 JainC#
// 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 Sam007PHP
$arr[$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
?>Javascript
输出 ：
Sum of maximum sum increasing subsequence is 106时间复杂度：O(n^2)
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