📌  相关文章
📜  最大化从矩阵中选择的K个元素的总和,以便每个选择的元素必须在选择的行元素之后

📅  最后修改于: 2021-05-18 00:24:41             🧑  作者: Mango

给定大小为N * M的2D数组arr [] []和整数K ,任务是选择具有最大可能和的K个元素,这样,如果选择了元素arr [i] [j] ,则所有元素从第i存在于j列的需求之前第i行被选择。

例子:

天真的方法:解决问题的最简单方法是根据指定的约束从给定的矩阵生成大小为K的所有可能子集,并从这些子集计算最大和。

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

高效方法:可以通过使用动态编程来优化上述方法。这个想法是找到选择i元素的最大和,直到2D数组arr [] []的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j行为。请按照以下步骤解决问题:

  • 初始化大小为(K + 1)*(N + 1)dp [] []表,并将dp [0] [i]初始化为0 ,因为没有元素的总和等于0
  • dp [i] [0]初始化为0 ,因为没有要选择的元素。
  • 使用两个嵌套循环在[0,K]范围内迭代,分别使用变量ij[0,N]范围内迭代:
    • 初始化一个变量,例如sum0 ,以跟踪arr [] []的j行中元素的累积和。
    • 如果不需要从arr [] []的j行中选择任何项,则将maxSum初始化为dp [i] [j ]
    • k1进行迭代,直到k> Mk> i为止:
      • 增量总和+ = buy [j] [k – 1]
      • 将现有的maxSum(sum + dp [i – k] [j])最大值存储maxSum中
    • dp [i] [j + 1]存储maxSum的值。
  • 打印值dp [K] [N]作为K个元素的最大和,直到矩阵arr [] []的(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


输出:
250

时间复杂度: O(K * N * M)
辅助空间: O(K * N)