通过从矩阵的不同部分中选择元素来最大化总和

给定一个N行M列的矩阵。假定M为3的倍数。列分为3个部分，第一部分为0到m / 3-1，第二部分为m / 3到2m / 3-1，第三部分为从2m / 3到m。任务是从每一行中选择一个元素，而在相邻行中，我们不能从同一部分中选择。我们必须最大化所选元素的总和。
例子：

Input: mat[][] = { 
{1, 3, 5, 2, 4, 6}, 
{6, 4, 5, 1, 3, 2}, 
{1, 3, 5, 2, 4, 6}, 
{6, 4, 5, 1, 3, 2}, 
{6, 4, 5, 1, 3, 2}, 
{1, 3, 5, 2, 4, 6}} 
Output: 35
Input: mat[][] = { 
{1, 2, 3}, 
{3, 2, 1}, 
{5, 4, 2} 
Output: 10 
 

方法：可以通过存储子问题并重用它们来使用动态编程解决方案来解决该问题。创建一个dp [] []数组，其中dp [i] [0]代表从0到i的行的元素总和，采用第1节的元素。类似地，对于dp [i] [1]和dp [i] [2] 。因此，打印max(dp [n – 1] [0]，dp [n – 1] [1]，dp [n – 1] [2] 。
下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
const int n = 6, m = 6;
 
// Function to find the maximum value
void maxSum(long arr[n][m])
{
    // Dp table
    long dp[n + 1][3] = { 0 };
 
    // Fill the dp in bottom
    // up manner
    for (int i = 0; i < n; i++) {
 
        // Maximum of the three sections
        long m1 = 0, m2 = 0, m3 = 0;
 
        for (int j = 0; j < m; j++) {
 
            // Maximum of the first section
            if ((j / (m / 3)) == 0) {
                m1 = max(m1, arr[i][j]);
            }
 
            // Maximum of the second section
            else if ((j / (m / 3)) == 1) {
                m2 = max(m2, arr[i][j]);
            }
 
            // Maximum of the third section
            else if ((j / (m / 3)) == 2) {
                m3 = max(m3, arr[i][j]);
            }
        }
 
        // If we choose element from section 1,
        // we cannot have selection from same section
        // in adjacent rows
        dp[i + 1][0] = max(dp[i][1], dp[i][2]) + m1;
        dp[i + 1][1] = max(dp[i][0], dp[i][2]) + m2;
        dp[i + 1][2] = max(dp[i][1], dp[i][0]) + m3;
    }
 
    // Print the maximum sum
    cout << max(max(dp[n][0], dp[n][1]), dp[n][2]) << '\n';
}
 
// Driver code
int main()
{
 
    long arr[n][m] = { { 1, 3, 5, 2, 4, 6 },
                       { 6, 4, 5, 1, 3, 2 },
                       { 1, 3, 5, 2, 4, 6 },
                       { 6, 4, 5, 1, 3, 2 },
                       { 6, 4, 5, 1, 3, 2 },
                       { 1, 3, 5, 2, 4, 6 } };
 
    maxSum(arr);
 
    return 0;
}

Java

// Java program for the above approach
class GFG
{
 
static int n = 6, m = 6;
 
// Function to find the maximum value
static void maxSum(long arr[][])
{
    // Dp table
    long [][]dp= new long[n + 1][3];
 
    // Fill the dp in bottom
    // up manner
    for (int i = 0; i < n; i++)
    {
 
        // Maximum of the three sections
        long m1 = 0, m2 = 0, m3 = 0;
 
        for (int j = 0; j < m; j++)
        {
 
            // Maximum of the first section
            if ((j / (m / 3)) == 0)
            {
                m1 = Math.max(m1, arr[i][j]);
            }
 
            // Maximum of the second section
            else if ((j / (m / 3)) == 1)
            {
                m2 = Math.max(m2, arr[i][j]);
            }
 
            // Maximum of the third section
            else if ((j / (m / 3)) == 2)
            {
                m3 = Math.max(m3, arr[i][j]);
            }
        }
 
        // If we choose element from section 1,
        // we cannot have selection from same section
        // in adjacent rows
        dp[i + 1][0] = Math.max(dp[i][1], dp[i][2]) + m1;
        dp[i + 1][1] = Math.max(dp[i][0], dp[i][2]) + m2;
        dp[i + 1][2] = Math.max(dp[i][1], dp[i][0]) + m3;
    }
 
    // Print the maximum sum
    System.out.print(Math.max(Math.max(dp[n][0], dp[n][1]), dp[n][2]) + "\n");
}
 
// Driver code
public static void main(String[] args)
{
 
    long arr[][] = { { 1, 3, 5, 2, 4, 6 },
                    { 6, 4, 5, 1, 3, 2 },
                    { 1, 3, 5, 2, 4, 6 },
                    { 6, 4, 5, 1, 3, 2 },
                    { 6, 4, 5, 1, 3, 2 },
                    { 1, 3, 5, 2, 4, 6 } };
 
    maxSum(arr);
}
}
 
// This code is contributed by PrinciRaj1992

Python3

# Python3 program for the above approach
import numpy as np
n = 6; m = 6;
 
# Function to find the maximum value
def maxSum(arr) :
 
    # Dp table
    dp = np.zeros((n + 1, 3));
 
    # Fill the dp in bottom
    # up manner
    for i in range(n) :
 
        # Maximum of the three sections
        m1 = 0; m2 = 0; m3 = 0;
         
        for j in range(m) :
             
            # Maximum of the first section
            if ((j // (m // 3)) == 0) :
                m1 = max(m1, arr[i][j]);
                 
            # Maximum of the second section
            elif ((j // (m // 3)) == 1) :
                m2 = max(m2, arr[i][j]);
                 
            # Maximum of the third section
            elif ((j // (m // 3)) == 2) :
                m3 = max(m3, arr[i][j]);
                 
        # If we choose element from section 1,
        # we cannot have selection from same section
        # in adjacent rows
        dp[i + 1][0] = max(dp[i][1], dp[i][2]) + m1;
        dp[i + 1][1] = max(dp[i][0], dp[i][2]) + m2;
        dp[i + 1][2] = max(dp[i][1], dp[i][0]) + m3;
 
    # Print the maximum sum
    print(max(max(dp[n][0], dp[n][1]), dp[n][2]));
 
# Driver code
if __name__ == "__main__" :
 
    arr = [[ 1, 3, 5, 2, 4, 6 ],
           [ 6, 4, 5, 1, 3, 2 ],
           [ 1, 3, 5, 2, 4, 6 ],
           [ 6, 4, 5, 1, 3, 2 ],
           [ 6, 4, 5, 1, 3, 2 ],
           [ 1, 3, 5, 2, 4, 6 ]];
 
    maxSum(arr);
     
# This code is contributed by AnkitRai01

C#

// C# program for the above approach
using System;
 
class GFG
{
static int n = 6, m = 6;
 
// Function to find the maximum value
static void maxSum(long [,]arr)
{
    // Dp table
    long [,]dp = new long[n + 1, 3];
 
    // Fill the dp in bottom
    // up manner
    for (int i = 0; i < n; i++)
    {
 
        // Maximum of the three sections
        long m1 = 0, m2 = 0, m3 = 0;
 
        for (int j = 0; j < m; j++)
        {
 
            // Maximum of the first section
            if ((j / (m / 3)) == 0)
            {
                m1 = Math.Max(m1, arr[i, j]);
            }
 
            // Maximum of the second section
            else if ((j / (m / 3)) == 1)
            {
                m2 = Math.Max(m2, arr[i, j]);
            }
 
            // Maximum of the third section
            else if ((j / (m / 3)) == 2)
            {
                m3 = Math.Max(m3, arr[i, j]);
            }
        }
 
        // If we choose element from section 1,
        // we cannot have selection from same section
        // in adjacent rows
        dp[i + 1, 0] = Math.Max(dp[i, 1], dp[i, 2]) + m1;
        dp[i + 1, 1] = Math.Max(dp[i, 0], dp[i, 2]) + m2;
        dp[i + 1, 2] = Math.Max(dp[i, 1], dp[i, 0]) + m3;
    }
 
    // Print the maximum sum
    Console.Write(Math.Max(Math.Max(dp[n, 0],
                                    dp[n, 1]),
                                    dp[n, 2]) + "\n");
}
 
// Driver code
public static void Main(String[] args)
{
    long [,]arr = { { 1, 3, 5, 2, 4, 6 },
                    { 6, 4, 5, 1, 3, 2 },
                    { 1, 3, 5, 2, 4, 6 },
                    { 6, 4, 5, 1, 3, 2 },
                    { 6, 4, 5, 1, 3, 2 },
                    { 1, 3, 5, 2, 4, 6 } };
 
    maxSum(arr);
}
}
 
// This code is contributed by 29AjayKumar

Javascript

// Javascript program for the above approach
 
const n = 6, m = 6;
 
// Function to find the maximum value
function maxSum(arr)
{
    // Dp table
    const dp = new Array(n+1).fill(0).map(() => new Array(3).fill(0));
 
    // Fill the dp in bottom
    // up manner
    for (var i = 0; i < n; i++) {
 
        // Maximum of the three sections
        var m1 = 0, m2 = 0, m3 = 0;
 
        for (var j = 0; j < m; j++) {
 
            // Maximum of the first section
            if (parseInt(j / (m / 3)) == 0) {
                m1 = Math.max(m1, arr[i][j]);
            }
 
            // Maximum of the second section
            else if (parseInt(j / (m / 3)) == 1) {
                m2 = Math.max(m2, arr[i][j]);
            }
 
            // Maximum of the third section
            else if (parseInt(j / (m / 3)) == 2) {
                m3 = Math.max(m3, arr[i][j]);
            }
        }
 
        // If we choose element from section 1,
        // we cannot have selection from same section
        // in adjacent rows
        dp[i + 1][0] = Math.max(dp[i][1], dp[i][2]) + m1;
        dp[i + 1][1] = Math.max(dp[i][0], dp[i][2]) + m2;
        dp[i + 1][2] = Math.max(dp[i][1], dp[i][0]) + m3;
    }
 
    // Print the maximum sum
    document.write( parseInt(Math.max(Math.max(dp[n][0], dp[n][1]), dp[n][2])) + "<br>");
}
 
arr = [ [ 1, 3, 5, 2, 4, 6 ],
        [ 6, 4, 5, 1, 3, 2 ],
        [ 1, 3, 5, 2, 4, 6 ],
        [ 6, 4, 5, 1, 3, 2 ],
        [ 6, 4, 5, 1, 3, 2 ],
        [ 1, 3, 5, 2, 4, 6 ] ];
 
maxSum(arr);
 
//This code is contributed by SoumikMondal

输出：

35

时间复杂度： O(N * M)

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