通过从矩阵的不同部分选择元素来最大化总和

给定一个N行M列的矩阵。假设M是 3 的倍数。 列被分为 3 段，第一段是从 0 到 m/3-1，第二段是从 m/3 到 2m/3-1，第三段是从 2m/3 到 m。任务是从每一行中选择一个元素，在相邻的行中，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：通过存储子问题并重用它们，可以使用动态规划解决方案来解决问题。创建一个dp[][]数组，其中dp[i][0]表示从0到i从第1部分获取元素的行的元素总和。同样，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)