最大化从矩阵中选择的 K 个元素的总和，以便每个选定的元素必须在选定的行元素之前

给定一个大小为N * M的二维数组arr[][]和一个整数K ，任务是选择具有最大可能总和的K 个元素，这样如果选择了一个元素arr[i][j] ，那么所有元素从第i存在于^第j列的需求之前^第i行被选择。

例子：

Input: arr[][] = {{10, 10, 100, 30}, {80, 50, 10, 50}}, K = 5
Output: 250
Explanation:
Selecting first 3 elements from the first row, sum = 10 + 10 + 100 = 120
Selecting first 2 elements from the second row, sum = 80 + 50 = 130
Therefore, the maximum sum = 120 + 130 = 250.

Input: arr[][] = {{30, 10, 110, 13}, {810, 152, 12, 5}, {124, 24, 54, 124}}, K = 6
Output: 1288
Explanation:
Selecting first 2 elements from the second row, sum = 810 + 152 = 962
Selecting all 4 elements from the third row, sum = 124 + 24 + 54 + 124 = 326
Therefore, the maximum sum = 962 + 326 = 1288

朴素方法：解决问题的最简单方法是根据指定的约束从给定矩阵生成大小为 K 的所有可能子集，并计算这些子集可能的最大和。

时间复杂度： O((N*M)!/(K!)*(N*M – K)!)
辅助空间： O(1)

高效方法：上述方法可以通过使用动态规划进行优化。这个想法是找到选择i元素的最大总和，直到二维数组arr[][] 的^第j行。辅助数组dp[i][j + 1]存储选择i 个元素的最大和，直到矩阵arr[][] 的^第j行。请按照以下步骤解决问题：

• 初始化大小为(K+1)*(N+1)的dp[][]表并将dp[0][i]初始化为0 ，因为没有元素的总和等于0 。
• 将dp[i][0]初始化为0 ，因为没有要选择的元素。
• 使用两个嵌套循环和[0, N] 分别使用变量i和j在范围[0, K] 上迭代：
  • 初始化一个变量，比如sum为0 ，以跟踪arr[][] 的^第j行元素的累积总和。
  • 如果不需要从arr[][] 的^第j行中选择项目，则将maxSum初始化为dp[i][j ] 。
  • 以k为1 进行迭代，直到k > M或k > i ：
    • 增加总和 += buy[j][k – 1] 。
    • 将现有maxSum和(sum + dp[i – k][j])的最大值存储在maxSum 中。
  • 将dp[i][j + 1] 存储为maxSum的值。
• 打印值dp[K][N]作为K 元素的最大总和，直到矩阵arr[][] 的^第(N – 1) 行。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to return the maximum
// of two elements
int max(int a, int b)
{
    return a > b ? a : b;
}
 
// Function to find the maximum sum
// of selecting K elements from
// the given 2D array arr[][]
int maximumsum(int arr[][4], int K,
               int N, int M)
{
    int sum = 0, maxSum;
    int i, j, k;
 
    // dp table of size (K+1)*(N+1)
    int dp[K + 1][N + 1];
 
    // Initialize dp[0][i] = 0
    for (i = 0; i <= N; i++)
        dp[0][i] = 0;
 
    // Initialize dp[i][0] = 0
    for (i = 0; i <= K; i++)
        dp[i][0] = 0;
 
    // Selecting i elements
    for (i = 1; i <= K; i++) {
 
        // Select i elements till jth row
        for (j = 0; j < N; j++) {
 
            // dp[i][j+1] is the maximum
            // of selecting i elements till
            // jth row
 
            // sum = 0, to keep track of
            // cummulative elements sum
            sum = 0;
            maxSum = dp[i][j];
 
            // Traverse arr[j][k] until
            // number of elements until k>i
            for (k = 1; k <= M && k <= i; k++) {
 
                // Select arr[j][k - 1]th item
                sum += arr[j][k - 1];
 
                maxSum
                    = max(maxSum,
                          sum + dp[i - k][j]);
            }
 
            // Store the maxSum in dp[i][j+1]
            dp[i][j + 1] = maxSum;
        }
    }
 
    // Return the maximum sum
    return dp[K][N];
}
 
// Driver Code
int main()
{
    int arr[][4] = { { 10, 10, 100, 30 },
                     { 80, 50, 10, 50 } };
 
    int N = 2, M = 4;
    int K = 5;
 
    // Function Call
    cout << maximumsum(arr, K, N, M);
 
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
    
class GFG{
    
// Function to return the maximum
// of two elements
static int max(int a, int b)
{
    return a > b ? a : b;
}
  
// Function to find the maximum sum
// of selecting K elements from
// the given 2D array arr[][]
static int maximumsum(int arr[][], int K,
                      int N, int M)
{
    int sum = 0, maxSum;
    int i, j, k;
  
    // dp table of size (K+1)*(N+1)
    int[][] dp = new int[K + 1][N + 1];
  
    // Initialize dp[0][i] = 0
    for(i = 0; i <= N; i++)
        dp[0][i] = 0;
  
    // Initialize dp[i][0] = 0
    for(i = 0; i <= K; i++)
        dp[i][0] = 0;
  
    // Selecting i elements
    for(i = 1; i <= K; i++)
    {
         
        // Select i elements till jth row
        for(j = 0; j < N; j++)
        {
             
            // dp[i][j+1] is the maximum
            // of selecting i elements till
            // jth row
  
            // sum = 0, to keep track of
            // cummulative elements sum
            sum = 0;
            maxSum = dp[i][j];
  
            // Traverse arr[j][k] until
            // number of elements until k>i
            for(k = 1; k <= M && k <= i; k++)
            {
                 
                // Select arr[j][k - 1]th item
                sum += arr[j][k - 1];
  
                maxSum = Math.max(maxSum,
                                  sum + dp[i - k][j]);
            }
  
            // Store the maxSum in dp[i][j+1]
            dp[i][j + 1] = maxSum;
        }
    }
  
    // Return the maximum sum
    return dp[K][N];
}
    
// Driver Code
public static void main(String[] args)
{
    int arr[][] =  { { 10, 10, 100, 30 },
                     { 80, 50, 10, 50 } };
  
    int N = 2, M = 4;
    int K = 5;
  
    // Function Call
    System.out.print(maximumsum(arr, K, N, M));
}
}
 
// This code is contributed by susmitakundugoaldanga

Python3

# Python program for the above approach
import math;
 
# Function to return the maximum
# of two elements
def max(a, b):
    if(a > b):
        return a;
    else:
        return b;
 
# Function to find the maximum sum
# of selecting K elements from
# the given 2D array arr
def maximumsum(arr, K, N, M):
    sum = 0;
    maxSum = 0;
 
    # dp table of size (K+1)*(N+1)
    dp = [[0 for i in range(N + 1)] for j in range(K + 1)]
 
    # Initialize dp[0][i] = 0
    for i in range(0, N + 1):
        dp[0][i] = 0;
 
    # Initialize dp[i][0] = 0
    for i in range(0, K + 1):
        dp[i][0] = 0;
 
    # Selecting i elements
    for  i in range(1, K + 1):
 
        # Select i elements till jth row
        for  j in range(0, N):
 
            # dp[i][j+1] is the maximum
            # of selecting i elements till
            # jth row
 
            # sum = 0, to keep track of
            # cummulative elements sum
            sum = 0;
            maxSum = dp[i][j];
 
            # Traverse arr[j][k] until
            # number of elements until k>i
            for k in range(1, i + 1):
                if(k > M):
                    break;
                     
                # Select arr[j][k - 1]th item
                sum += arr[j][k - 1];
 
                maxSum = max(maxSum, sum + dp[i - k][j]);
 
            # Store the maxSum in dp[i][j+1]
            dp[i][j + 1] = maxSum;
 
    # Return the maximum sum
    return dp[K][N];
 
# Driver Code
if __name__ == '__main__':
    arr = [[10, 10, 100, 30], [80, 50, 10, 50]];
 
    N = 2;
    M = 4;
    K = 5;
 
    # Function Call
    print(maximumsum(arr, K, N, M));
 
 
    # This code is contributed by 29AjayKumar

C#

// C# program for the above approach
using System;
 
class GFG{
    
// Function to return the maximum
// of two elements
static int max(int a, int b)
{
    return a > b ? a : b;
}
  
// Function to find the maximum sum
// of selecting K elements from
// the given 2D array [,]arr
static int maximumsum(int [,]arr, int K,
                      int N, int M)
{
    int sum = 0, maxSum;
    int i, j, k;
     
    // dp table of size (K+1)*(N+1)
    int[,] dp = new int[K + 1, N + 1];
  
    // Initialize dp[0,i] = 0
    for(i = 0; i <= N; i++)
        dp[0, i] = 0;
  
    // Initialize dp[i,0] = 0
    for(i = 0; i <= K; i++)
        dp[i, 0] = 0;
  
    // Selecting i elements
    for(i = 1; i <= K; i++)
    {
         
        // Select i elements till jth row
        for(j = 0; j < N; j++)
        {
             
            // dp[i,j+1] is the maximum
            // of selecting i elements till
            // jth row
  
            // sum = 0, to keep track of
            // cummulative elements sum
            sum = 0;
            maxSum = dp[i, j];
  
            // Traverse arr[j,k] until
            // number of elements until k>i
            for(k = 1; k <= M && k <= i; k++)
            {
                 
                // Select arr[j,k - 1]th item
                sum += arr[j, k - 1];
  
                maxSum = Math.Max(maxSum,
                                  sum + dp[i - k, j]);
            }
  
            // Store the maxSum in dp[i,j+1]
            dp[i, j + 1] = maxSum;
        }
    }
  
    // Return the maximum sum
    return dp[K, N];
}
    
// Driver Code
public static void Main(String[] args)
{
    int [,]arr =  { { 10, 10, 100, 30 },
                    { 80, 50, 10, 50 } };
  
    int N = 2, M = 4;
    int K = 5;
  
    // Function Call
    Console.Write(maximumsum(arr, K, N, M));
}
}
 
// This code is contributed by Amit Katiyar

Javascript

// Javascript program to implement
// the above approach
 
// Function to return the maximum
// of two elements
function max(a, b)
{
    return a > b ? a : b;
}
   
// Function to find the maximum sum
// of selecting K elements from
// the given 2D array arr[][]
function maximumsum(arr, K,
                      N, M)
{
    let sum = 0, maxSum;
    let i, j, k;
   
    // dp table of size (K+1)*(N+1)
    let dp = new Array(K + 1);
     
    // Loop to create 2D array using 1D array
    for ( i = 0; i < dp.length; i++) {
        dp[i] = new Array(2);
    }
   
    // Initialize dp[0][i] = 0
    for(i = 0; i <= N; i++)
        dp[0][i] = 0;
   
    // Initialize dp[i][0] = 0
    for(i = 0; i <= K; i++)
        dp[i][0] = 0;
   
    // Selecting i elements
    for(i = 1; i <= K; i++)
    {
          
        // Select i elements till jth row
        for(j = 0; j < N; j++)
        {
              
            // dp[i][j+1] is the maximum
            // of selecting i elements till
            // jth row
   
            // sum = 0, to keep track of
            // cummulative elements sum
            sum = 0;
            maxSum = dp[i][j];
   
            // Traverse arr[j][k] until
            // number of elements until k>i
            for(k = 1; k <= M && k <= i; k++)
            {
                  
                // Select arr[j][k - 1]th item
                sum += arr[j][k - 1];
   
                maxSum = Math.max(maxSum,
                                  sum + dp[i - k][j]);
            }
   
            // Store the maxSum in dp[i][j+1]
            dp[i][j + 1] = maxSum;
        }
    }
   
    // Return the maximum sum
    return dp[K][N];
}
 
// Driver code
    let arr =  [[ 10, 10, 100, 30 ],
                     [ 80, 50, 10, 50 ]];
   
    let N = 2, M = 4;
    let K = 5;
   
    // Function Call
    document.write(maximumsum(arr, K, N, M));
  
 // This code is contributed by souravghosh0416.

输出：

250

时间复杂度： O(K*N*M)
辅助空间： O(K*N)

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