通过从给定矩阵的每个单元格对角遍历来最大化总和

给定一个维度为N x N的二维方阵arr[][] ，任务是通过从任何单元格对角移动来找到最大路径和，并且每个单元格必须仅访问一次，即从单元格(i, j) ，玩家可以移动到单元格(i + 1, j + 1) 。

例子：

Input: arr[][] = {{1, 2, 3}, {3, 5, 10}, {1 3 5}} 
Output: 12
Explanation:
Sum of cells (1, 1), (2, 2) and (3, 3) is 11.  
The sum of cells (1, 2), (2, 3) and (1, 3) is 3. 
The sum of cells (2, 1) and (3, 2) is 6.
The sum of cell (3, 1) is 1.
The maximum possible sum is 12.

Input: arr[][] = {{1, 1, 1}, {1 1 1}, {1 1 1}} 
Output: 3

方法：解决这个问题的思路是对第一行第一列元素对角遍历矩阵，在矩阵范围内求和它们的对角元素。
请按照以下步骤解决问题：

1. 初始化一个变量，比如max为0 。
2. 从第一行和第一列中选择每个单元格(i, j) 。
3. 现在，从每个单元格中，通过将i和j增加1 来找到从该单元格开始的对角线和，例如sum 。
4. 然后，将max更新为max(max, sum) 。
5. 遍历后，打印max作为所需的答案。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to find the maximum sum
int MaximumSum(vector >& arr, int n)
{
 
    int ans = 0;
 
    // Loop to traverse through the
    // upper triangular matrix and
    // update the maximum sum to ans
    for (int i = 0; i < n; i++) {
        int x = 0, y = i, sum = 0;
        for (int j = i; j < n; j++) {
            sum += arr[x++][y++];
        }
        if (sum > ans)
            ans = sum;
    }
 
    // Traverse through the
    // lower triangular matrix
    for (int i = 1; i < n; i++) {
 
        int x = i, y = 0, sum = 0;
 
        for (int j = i; j < n; j++) {
 
            sum += arr[x++][y++];
        }
        if (sum > ans)
            ans = sum;
    }
 
    return ans;
}
 
// Driver Code
int main()
{
 
    // Given matrix
    vector > arr;
    arr = { { 1, 2, 3 },
            { 3, 5, 10 },
            { 1, 3, 5 } };
 
    // Given dimension
    int n = arr.size();
 
    cout << MaximumSum(arr, n);
 
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
class GFG{
 
// Function to find the maximum sum
static int MaximumSum(int [][]arr, int n)
{
 
    int ans = 0;
 
    // Loop to traverse through the
    // upper triangular matrix and
    // update the maximum sum to ans
    for (int i = 0; i < n; i++)
    {
        int x = 0, y = i, sum = 0;
        for (int j = i; j < n; j++)
        {
            sum += arr[x++][y++];
        }
        if (sum > ans)
            ans = sum;
    }
 
    // Traverse through the
    // lower triangular matrix
    for (int i = 1; i < n; i++)
    {
        int x = i, y = 0, sum = 0;
        for (int j = i; j < n; j++)
        {
            sum += arr[x++][y++];
        }
        if (sum > ans)
            ans = sum;
    }
    return ans;
}
 
// Driver Code
public static void main(String[] args)
{
 
    // Given matrix
    int [][]arr = { { 1, 2, 3 },
            { 3, 5, 10 },
            { 1, 3, 5 } };
 
    // Given dimension
    int n = arr.length;
    System.out.print(MaximumSum(arr, n));
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 program for the above approach
 
# Function to find the maximum sum
def MaximumSum(arr, n):
    ans = 0;
 
    # Loop to traverse through the
    # upper triangular matrix and
    # update the maximum sum to ans
    for i in range(n):
        x, y, sum = 0, i, 0
        for j in range(i, n):
            sum, x, y =sum + arr[x][y], x + 1, y + 1
        if (sum > ans):
            ans = sum
 
    # Traverse through the
    # lower triangular matrix
    for i in range(1, n):
 
        x, y, sum = i, 0, 0
 
        for j in range(i, n):
 
            sum, x, y =sum + arr[x][y], x + 1, y + 1
        if (sum > ans):
            ans = sum
    return ans
 
# Driver Code
if __name__ == '__main__':
 
    # Given matrix
    arr = [ [ 1, 2, 3],
            [ 3, 5, 10],
            [ 1, 3, 5 ]]
 
    # Given dimension
    n = len(arr)
    print (MaximumSum(arr, n))
 
    # This code is contributed by mohit kumar 29

C#

// C# program for the above approach
using System;
class GFG{
 
  // Function to find the maximum sum
  static int MaximumSum(int [,]arr, int n)
  {
    int ans = 0;
 
    // Loop to traverse through the
    // upper triangular matrix and
    // update the maximum sum to ans
    for (int i = 0; i < n; i++)
    {
      int x = 0, y = i, sum = 0;
      for (int j = i; j < n; j++)
      {
        sum += arr[x++, y++];
      }
      if (sum > ans)
        ans = sum;
    }
 
    // Traverse through the
    // lower triangular matrix
    for (int i = 1; i < n; i++)
    {
      int x = i, y = 0, sum = 0;
      for (int j = i; j < n; j++)
      {
        sum += arr[x++, y++];
      }
      if (sum > ans)
        ans = sum;
    }
    return ans;
  }
 
  // Driver Code
  public static void Main(String[] args)
  {
 
    // Given matrix
    int [,]arr = { { 1, 2, 3 },
                  { 3, 5, 10 },
                  { 1, 3, 5 } };
 
    // Given dimension
    int n = arr.GetLength(0);
    Console.Write(MaximumSum(arr, n));
  }
}
 
// This code is contributed by shikhasingrajput

Javascript

 
// Javascript program of the above approach
 
// Function to find the maximum sum
function MaximumSum(arr, n)
{
  
    let ans = 0;
  
    // Loop to traverse through the
    // upper triangular matrix and
    // update the maximum sum to ans
    for (let i = 0; i < n; i++)
    {
        let x = 0, y = i, sum = 0;
        for (let j = i; j < n; j++)
        {
            sum += arr[x++][y++];
        }
        if (sum > ans)
            ans = sum;
    }
  
    // Traverse through the
    // lower triangular matrix
    for (let i = 1; i < n; i++)
    {
        let x = i, y = 0, sum = 0;
        for (let j = i; j < n; j++)
        {
            sum += arr[x++][y++];
        }
        if (sum > ans)
            ans = sum;
    }
    return ans;
}
 
    // Driver Code
     
    // Given matrix
    let arr = [[ 1, 2, 3 ],
            [ 3, 5, 10 ],
            [ 1, 3, 5 ]];
  
    // Given dimension
    let n = arr.length;
    document.write(MaximumSum(arr, n));
  

输出：

12

时间复杂度： O(N ² )
辅助空间： O(N ² )

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