来自 n 个数组的递增顺序元素的最大总和

给定 n 个大小为 m 的数组。求通过从每个数组中选择一个数字获得的最大和，使得从第 i 个数组中选择的元素多于从第 (i-1) 个数组中选择的元素。如果无法获得最大和，则返回 0。
例子：

Input : arr[][] = {{1, 7, 3, 4},
                   {4, 2, 5, 1},
                   {9, 5, 1, 8}}
Output : 18
Explanation :
We can select 4 from first array, 5 from 
second array and 9 from third array.

Input : arr[][] = {{9, 8, 7},
                   {6, 5, 4},
                   {3, 2, 1}}
Output : 0

这个想法是从最后一个数组开始挑选。我们从最后一个数组中选取最大的元素，然后移动到倒数第二个数组。在倒数第二个数组中，我们找到小于从最后一个数组中选取的最大元素的最大元素。我们重复这个过程，直到我们到达第一个数组。
为了获得最大和，我们可以对所有数组进行排序，然后从下到上从右到左遍历每个数组，然后选择一个大于前一个元素的数字。如果我们无法从数组中选择一个元素，则返回 0。

C++

// CPP program to find maximum sum
// by selecting a element from n arrays
#include 
#define M 4
using namespace std;
 
// To calculate maximum sum by
// selecting element from each array
int maximumSum(int a[][M], int n) {
 
  // Sort each array
  for (int i = 0; i < n; i++)
    sort(a[i], a[i] + M);
 
  // Store maximum element
  // of last array
  int sum = a[n - 1][M - 1];
  int prev = a[n - 1][M - 1];
  int i, j;
 
  // Selecting maximum element from
  // previoulsy selected element
  for (i = n - 2; i >= 0; i--) {
    for (j = M - 1; j >= 0; j--) {
      if (a[i][j] < prev) {
        prev = a[i][j];
        sum += prev;
        break;
      }
    }
 
    // j = -1 means no element is
    // found in a[i] so return 0
    if (j == -1)
      return 0;
  }
 
  return sum;
}
 
// Driver program to test maximumSum
int main() {
  int arr[][M] = {{1, 7, 3, 4},
                  {4, 2, 5, 1},
                  {9, 5, 1, 8}};
  int n = sizeof(arr) / sizeof(arr[0]);
  cout << maximumSum(arr, n);
  return 0;
}

Java

// Java program to find
// maximum sum by selecting
// a element from n arrays
import java.io.*;
 
class GFG
{
    static int M = 4;
    static int arr[][] = {{1, 7, 3, 4},
                          {4, 2, 5, 1},
                          {9, 5, 1, 8}};
 
    static void sort(int a[][],
                     int row, int n)
    {
        for (int i = 0; i < M - 1; i++)
        {
            if(a[row][i] > a[row][i + 1])
            {
                int temp = a[row][i];
                a[row][i] = a[row][i + 1];
                a[row][i + 1] = temp;
            }
        }
    }
     
    // To calculate maximum
    // sum by selecting element
    // from each array
    static int maximumSum(int a[][],
                          int n)
    {
     
    // Sort each array
    for (int i = 0; i < n; i++)
        sort(a, i, n);
     
    // Store maximum element
    // of last array
    int sum = a[n - 1][M - 1];
    int prev = a[n - 1][M - 1];
    int i, j;
     
    // Selecting maximum element
    // from previoulsy selected
    // element
    for (i = n - 2; i >= 0; i--)
    {
        for (j = M - 1; j >= 0; j--)
        {
            if (a[i][j] < prev)
            {
                prev = a[i][j];
                sum += prev;
                break;
            }
        }
     
        // j = -1 means no element
        // is found in a[i] so
        // return 0
        if (j == -1)
        return 0;
    }    
    return sum;
    }
     
    // Driver Code
    public static void main(String args[])
    {
        int n = arr.length;
        System.out.print(maximumSum(arr, n));
    }
}
 
// This code is contributed by
// Manish Shaw(manishshaw1)

Python3

# Python3 program to find
# maximum sum by selecting
# a element from n arrays
M = 4;
 
# To calculate maximum sum
# by selecting element from
# each array
def maximumSum(a, n) :
 
    global M;
     
    # Sort each array
    for i in range(0, n) :
        a[i].sort();
     
    # Store maximum element
    # of last array
    sum = a[n - 1][M - 1];
    prev = a[n - 1][M - 1];
 
    # Selecting maximum
    # element from previoulsy
    # selected element
    for i in range(n - 2,
                  -1, -1) :
     
        for j in range(M - 1,
                      -1, -1) :
         
            if (a[i][j] < prev) :
             
                prev = a[i][j];
                sum += prev;
                break;
 
        # j = -1 means no element
        # is found in a[i] so
        # return 0
        if (j == -1) :
            return 0;
    return sum;
 
# Driver Code
arr = [[1, 7, 3, 4],
       [4, 2, 5, 1],
       [9, 5, 1, 8]];
n = len(arr) ;
print (maximumSum(arr, n));
 
# This code is contributed by
# Manish Shaw(manishshaw1)

C#

// C# program to find maximum
// sum by selecting a element
// from n arrays
using System;
 
class GFG
{
    static int M = 4;
     
    static void sort(ref int[,] a,
                     int row, int n)
    {
        for (int i = 0; i < M-1; i++)
        {
            if(a[row, i] > a[row, i + 1])
            {
                int temp = a[row, i];
                a[row, i] = a[row, i + 1];
                a[row, i + 1] = temp;
            }
        }
    }
     
    // To calculate maximum
    // sum by selecting
    // element from each array
    static int maximumSum(int[,] a,
                          int n)
    {
     
    int i = 0, j = 0;
     
    // Sort each array
    for (i = 0; i < n; i++)
        sort(ref a, i, n);
     
    // Store maximum element
    // of last array
    int sum = a[n - 1, M - 1];
    int prev = a[n - 1, M - 1];
     
     
    // Selecting maximum element
    // from previoulsy selected
    // element
    for (i = n - 2; i >= 0; i--)
    {
        for (j = M - 1; j >= 0; j--)
        {
        if (a[i, j] < prev)
        {
            prev = a[i, j];
            sum += prev;
            break;
        }
        }
     
        // j = -1 means no element
        // is found in a[i] so
        // return 0
        if (j == -1)
        return 0;
    }
     
    return sum;
    }
     
    // Driver Code
    static void Main()
    {
    int [,]arr = new int[,]{{1, 7, 3, 4},
                            {4, 2, 5, 1},
                            {9, 5, 1, 8}};
    int n = arr.GetLength(0);
    Console.Write(maximumSum(arr, n));
    }
}
 
// This code is contributed by
// Manish Shaw (manishshaw1)

PHP

= 0; $i--)
    {
        for ($j = $M - 1; $j >= 0; $j--)
        {
        if ($a[$i][$j] < $prev)
        {
            $prev = $a[$i][$j];
            $sum += $prev;
            break;
        }
        }
     
        // j = -1 means no element is
        // found in a[i] so return 0
        if ($j == -1)
        return 0;
    }
     
    return $sum;
}
 
// Driver Code
$arr = array(array(1, 7, 3, 4),
             array(4, 2, 5, 1),
             array(9, 5, 1, 8));
$n = sizeof($arr) ;
echo maximumSum($arr, $n);
 
// This code is contributed by m_kit
?>

Javascript

 
// JavaScript program to find
// maximum sum by selecting
// a element from n arrays
 
let M = 4;
  
// To calculate maximum sum
// by selecting element from
// each array
function maximumSum(a, n)
{
  // Store maximum element
  // of last array
  let prev = Math.max(a[n - 1][0],
                      a[n - 1][M - 1] + 1);
  
  // Selecting maximum element from
  // previoulsy selected element
  let sum = prev;
  for (let i = n - 2; i >= 0; i--)
  {
    let max_smaller = Number.MIN_VALUE;
    for (let j = M - 1; j >= 0; j--)
    {
      if (a[i][j] < prev &&
          a[i][j] > max_smaller)
        max_smaller = a[i][j];
    }
  
    // max_smaller equals to
    // INT_MIN means no element
    // is found in a[i] so return 0
    if (max_smaller == Number.MIN_VALUE)
      return 0;
  
    prev = max_smaller;
    sum += max_smaller;
  }
  
  return sum;
}
 
// Driver code
 
        let arr = [[1, 7, 3, 4],
                 [4, 2, 5, 1],
                 [9, 5, 1, 8]];
  let n = arr.length;
  document.write(maximumSum(arr, n));
 

输出：

18

最坏情况时间复杂度：O(mn Log m)
我们可以优化上述解决方案以在 O(mn) 中工作。我们可以跳过排序来查找最大元素。

C++

// CPP program to find maximum sum
// by selecting a element from n arrays
#include 
#define M 4
using namespace std;
 
// To calculate maximum sum by
// selecting element from each array
int maximumSum(int a[][M], int n) {
 
  // Store maximum element of last array
  int prev = *max_element(&a[n-1][0],
                   &a[n-1][M-1] + 1);
 
  // Selecting maximum element from
  // previoulsy selected element
  int sum = prev;
  for (int i = n - 2; i >= 0; i--) {
 
    int max_smaller = INT_MIN;
    for (int j = M - 1; j >= 0; j--) {
      if (a[i][j] < prev &&
          a[i][j] > max_smaller)
        max_smaller = a[i][j];
    }
 
    // max_smaller equals to INT_MIN means
    // no element is found in a[i] so
    // return 0
    if (max_smaller == INT_MIN)
      return 0;
 
    prev = max_smaller;
    sum += max_smaller;
  }
 
  return sum;
}
 
// Driver program to test maximumSum
int main() {
  int arr[][M] = {{1, 7, 3, 4},
                  {4, 2, 5, 1},
                  {9, 5, 1, 8}};
  int n = sizeof(arr) / sizeof(arr[0]);
  cout << maximumSum(arr, n);
  return 0;
}

Java

// Java program to find
// maximum sum by selecting
// a element from n arrays
import java.util.*;
class GFG{
  
static int M = 4;
 
// To calculate maximum sum
// by selecting element from
// each array
static int maximumSum(int a[][],
                      int n)
{
  // Store maximum element
  // of last array
  int prev = Math.max(a[n - 1][0],
                      a[n - 1][M - 1] + 1);
 
  // Selecting maximum element from
  // previoulsy selected element
  int sum = prev;
  for (int i = n - 2; i >= 0; i--)
  {
    int max_smaller = Integer.MIN_VALUE;
    for (int j = M - 1; j >= 0; j--)
    {
      if (a[i][j] < prev &&
          a[i][j] > max_smaller)
        max_smaller = a[i][j];
    }
 
    // max_smaller equals to
    // INT_MIN means no element
    // is found in a[i] so return 0
    if (max_smaller == Integer.MIN_VALUE)
      return 0;
 
    prev = max_smaller;
    sum += max_smaller;
  }
 
  return sum;
}
 
// Driver code
public static void main(String []args)
{
  int arr[][] = {{1, 7, 3, 4},
                 {4, 2, 5, 1},
                 {9, 5, 1, 8}};
  int n = arr.length;
  System.out.print(maximumSum(arr, n)); 
}
}
 
// This code is contributed by Chitranayal

蟒蛇3

# Python3 program to find maximum sum
# by selecting a element from n arrays
M = 4
 
# To calculate maximum sum by
# selecting element from each array
def maximumSum(a, n):
 
    # Store maximum element of last array
    prev = max(max(a))
 
    # Selecting maximum element from
    # previoulsy selected element
    Sum = prev
    for i in range(n - 2, -1, -1):
 
        max_smaller = -10**9
        for j in range(M - 1, -1, -1):
            if (a[i][j] < prev and
                a[i][j] > max_smaller):
                max_smaller = a[i][j]
 
    # max_smaller equals to INT_MIN means
    # no element is found in a[i] so
    # return 0
        if (max_smaller == -10**9):
            return 0
 
        prev = max_smaller
        Sum += max_smaller
 
    return Sum
 
# Driver Code
arr = [[1, 7, 3, 4],
       [4, 2, 5, 1],
       [9, 5, 1, 8]]
n = len(arr)
print(maximumSum(arr, n))
 
# This code is contributed by mohit kumar

C＃

// C# program to find
// maximum sum by selecting
// a element from n arrays
using System;
class GFG
{
  static int M = 4;
 
  // To calculate maximum sum
  // by selecting element from
  // each array
  static int maximumSum(int[,] a, int n)
  {
 
    // Store maximum element
    // of last array
    int prev = Math.Max(a[n - 1, 0],
                        a[n - 1, M - 1] + 1);
 
    // Selecting maximum element from
    // previoulsy selected element
    int sum = prev;
    for(int i = n - 2; i >= 0; i--)
    {
      int max_smaller = Int32.MinValue;
      for(int j = M - 1; j >= 0; j--)
      {
        if(a[i, j] < prev && a[i, j] > max_smaller)
        {
          max_smaller = a[i, j];
        }
      }
 
      // max_smaller equals to
      // INT_MIN means no element
      // is found in a[i] so return 0
      if(max_smaller == Int32.MinValue)
      {
        return 0;
      }
      prev = max_smaller;
      sum += max_smaller;
    }
    return sum;
  }
 
  // Driver code
  static public void Main ()
  {
    int[,] arr = {{1, 7, 3, 4},{4, 2, 5, 1},{9, 5, 1, 8}};
    int n = arr.GetLength(0);
    Console.Write(maximumSum(arr, n));
  }
}
 
// This code is contributed by avanitrachhadiya2155

Javascript

 
// JavaScript program to find
// maximum sum by selecting
// a element from n arrays
 
let M = 4;
  
// To calculate maximum sum
// by selecting element from
// each array
function maximumSum(a, n)
{
  // Store maximum element
  // of last array
  let prev = Math.max(a[n - 1][0],
                      a[n - 1][M - 1] + 1);
  
  // Selecting maximum element from
  // previoulsy selected element
  let sum = prev;
  for (let i = n - 2; i >= 0; i--)
  {
    let max_smaller = Number.MIN_VALUE;
    for (let j = M - 1; j >= 0; j--)
    {
      if (a[i][j] < prev &&
          a[i][j] > max_smaller)
        max_smaller = a[i][j];
    }
  
    // max_smaller equals to
    // INT_MIN means no element
    // is found in a[i] so return 0
    if (max_smaller == Number.MIN_VALUE)
      return 0;
  
    prev = max_smaller;
    sum += max_smaller;
  }
  
  return sum;
}
 
// Driver code
 
        let arr = [[1, 7, 3, 4],
                 [4, 2, 5, 1],
                 [9, 5, 1, 8]];
  let n = arr.length;
  document.write(maximumSum(arr, n));
 
// This code is contributed by souravghosh0416.

输出：

18

时间复杂度： O(mn)

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