按行和列排序的矩阵的任何子矩阵的最大和

给定一个矩阵mat[][] ，其元素按行和列排序。任务是从给定的矩阵mat[][] 中找到任何子矩阵的最大和。

例子：

Input: mat[][] = { {-6, -4, -1}, {-3, 2, 4}, {2, 5, 8}} 
Output: 19 
Explanation: 
The largest submatrix is given by: 
2 4 
5 8
Input: mat[][] = { {-4, -3}, {-2, -1} } 
Output: -1 
Explanation: 
The sub matrix consisting of the last element i.e., -1 has the largest sum possible. 
 

朴素的方法：这个想法是使用 Kadane 算法在 2D 矩阵中找到最大和矩形。打印获得的最大和。

时间复杂度： O(N ² *M ² )，其中 N 是行数，M 是列数
辅助空间： O(N)
有效方法：思想是从给定矩阵的底部单元格中找到最大和，并使用动态规划来存储来自底部单元格的任何子矩阵的最大和。以下是步骤：

1. 创建一个大小为NxM的 dp 表dp[][]来存储从每个单元格(i, j)开始的子矩阵的最大和。
2. 找到从右下角的单元格(N, M)开始向上和向左的子矩阵的总和，并不断将最大总和更新为dp[][] 。
3. 由于矩阵按 Row-Wise 和 Column-Wise 排序，因此最大的子矩阵总和可以从任何点开始，但肯定会在右下角的单元格 (N, M) 处结束。
4. 下面是 dp 表的填充关系：

DP[i][j] = DP[i+1][j] + DP[i][j+1] – DP[i+1][j+1] 
 

1. 因此，在 dp 表中找到最大元素。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function that finds the maximum
// Sub-Matrix Sum
int maxSubMatSum(vector > mat)
{
    // Number of rows in the matrix
    int n = mat.size();
 
    // Number of columns in the matrix
    int m = mat[0].size();
 
    int i, j;
 
    // dp[][] matrix to store the
    // results of each iteration
    int dp[n][m];
 
    // Base Case - The largest
    // element in the matrix
    dp[n - 1][m - 1] = mat[n - 1][m - 1];
 
    // To stores the final result
    int res = dp[n - 1][m - 1];
 
    // Find the max sub matrix sum for
    // the last row
    for (i = m - 2; i >= 0; i--) {
 
        dp[n - 1][i] = mat[n - 1][i]
                       + dp[n - 1][i + 1];
 
        // Check whether the current
        // sub-array yeilds maximum sum
        res = max(res, dp[n - 1][i]);
    }
 
    // Calculate the max sub matrix
    // sum for the last column
    for (i = n - 2; i >= 0; i--) {
 
        dp[i][m - 1] = mat[i][m - 1]
                       + dp[i + 1][m - 1];
 
        // Check whether the current
        // sub-array yeilds maximum sum
        res = max(res, dp[i][m - 1]);
    }
 
    // Build the dp[][] matrix from
    // bottom to the top row
    for (i = n - 2; i >= 0; i--) {
 
        for (j = m - 2; j >= 0; j--) {
 
            // Update sum at each
            // cell in dp[][]
            dp[i][j]
                = mat[i][j] + dp[i][j + 1]
                  + dp[i + 1][j]
                  - dp[i + 1][j + 1];
 
            // Update the maximum sum
            res = max(res, dp[i][j]);
        }
    }
 
    // Return the maximum sum
    return res;
}
 
// Driver Code
int main()
{
    // Given matrix mat[][]
    vector > mat;
    mat = { { -6, -4, -1 },
            { -3, 2, 4 },
            { 2, 5, 8 } };
 
    // Function Call
    cout << maxSubMatSum(mat);
    return 0;
}

Java

// Java program for the above approach
class GFG{
 
// Function that finds the maximum
// Sub-Matrix Sum
static int maxSubMatSum(int [][]mat)
{
    // Number of rows in the matrix
    int n = mat.length;
 
    // Number of columns in the matrix
    int m = mat[0].length;
 
    int i, j;
 
    // dp[][] matrix to store the
    // results of each iteration
    int [][]dp = new int[n][m];
 
    // Base Case - The largest
    // element in the matrix
    dp[n - 1][m - 1] = mat[n - 1][m - 1];
 
    // To stores the final result
    int res = dp[n - 1][m - 1];
 
    // Find the max sub matrix sum for
    // the last row
    for (i = m - 2; i >= 0; i--)
    {
        dp[n - 1][i] = mat[n - 1][i] +
                        dp[n - 1][i + 1];
 
        // Check whether the current
        // sub-array yeilds maximum sum
        res = Math.max(res, dp[n - 1][i]);
    }
 
    // Calculate the max sub matrix
    // sum for the last column
    for (i = n - 2; i >= 0; i--)
    {
        dp[i][m - 1] = mat[i][m - 1] +
                        dp[i + 1][m - 1];
 
        // Check whether the current
        // sub-array yeilds maximum sum
        res = Math.max(res, dp[i][m - 1]);
    }
 
    // Build the dp[][] matrix from
    // bottom to the top row
    for (i = n - 2; i >= 0; i--)
    {
        for (j = m - 2; j >= 0; j--)
        {
 
            // Update sum at each
            // cell in dp[][]
            dp[i][j] = mat[i][j] + dp[i][j + 1] +
                    dp[i + 1][j] - dp[i + 1][j + 1];
 
            // Update the maximum sum
            res = Math.max(res, dp[i][j]);
        }
    }
 
    // Return the maximum sum
    return res;
}
 
// Driver Code
public static void main(String[] args)
{
    // Given matrix mat[][]
    int [][]mat= {{ -6, -4, -1 },
                  { -3, 2, 4 },
                  { 2, 5, 8 } };
 
    // Function Call
    System.out.print(maxSubMatSum(mat));
}
}
 
// This code is contributed by Rohit_ranjan

Python3

# Python3 program for the above approach
 
# Function that finds the maximum
# Sub-Matrix Sum
def maxSubMatSum(mat):
     
    # Number of rows in the matrix
    n = len(mat)
 
    # Number of columns in the matrix
    m = len(mat[0])
 
    # dp[][] matrix to store the
    # results of each iteration
    dp = [[0] * m for _ in range(n)]
 
    # Base Case - The largest
    # element in the matrix
    dp[n - 1][m - 1] = mat[n - 1][m - 1]
 
    # To stores the final result
    res = dp[n - 1][m - 1]
 
    # Find the max sub matrix sum for
    # the last row
    for i in range(m - 2, -1, -1):
        dp[n - 1][i] = (mat[n - 1][i] +
                         dp[n - 1][i + 1])
 
        # Check whether the current
        # sub-array yeilds maximum sum
        res = max(res, dp[n - 1][i])
 
    # Calculate the max sub matrix
    # sum for the last column
    for i in range(n - 2, -1, -1):
        dp[i][m - 1] = (mat[i][m - 1] +
                     dp[i + 1][m - 1])
 
        # Check whether the current
        # sub-array yeilds maximum sum
        res = max(res, dp[i][m - 1])
 
    # Build the dp[][] matrix from
    # bottom to the top row
    for i in range(n - 2, -1, -1):
        for j in range(m - 2, -1, -1):
 
            # Update sum at each
            # cell in dp[][]
            dp[i][j] = (mat[i][j] +
                         dp[i][j + 1] +
                         dp[i + 1][j]-
                         dp[i + 1][j + 1])
 
            # Update the maximum sum
            res = max(res, dp[i][j])
 
    # Return the maximum sum
    return res
 
# Driver Code
if __name__ == '__main__':
     
    # Given matrix mat[][]
    mat = [ [ -6, -4, -1 ],
            [ -3, 2, 4 ],
            [ 2, 5, 8 ] ]
 
    # Function call
    print(maxSubMatSum(mat))
 
# This code is contributed by mohit kumar 29

C#

// C# program for the above approach
using System;
class GFG{
 
// Function that finds the maximum
// Sub-Matrix Sum
static int maxSubMatSum(int [,]mat)
{
    // Number of rows in the matrix
    int n = mat.GetLength(0);
 
    // Number of columns in the matrix
    int m = mat.GetLength(1);
 
    int i, j;
 
    // [,]dp matrix to store the
    // results of each iteration
    int [,]dp = new int[n, m];
 
    // Base Case - The largest
    // element in the matrix
    dp[n - 1, m - 1] = mat[n - 1, m - 1];
 
    // To stores the readonly result
    int res = dp[n - 1, m - 1];
 
    // Find the max sub matrix sum for
    // the last row
    for (i = m - 2; i >= 0; i--)
    {
        dp[n - 1, i] = mat[n - 1, i] +
                        dp[n - 1, i + 1];
 
        // Check whether the current
        // sub-array yeilds maximum sum
        res = Math.Max(res, dp[n - 1,i]);
    }
 
    // Calculate the max sub matrix
    // sum for the last column
    for (i = n - 2; i >= 0; i--)
    {
        dp[i, m - 1] = mat[i, m - 1] +
                        dp[i + 1, m - 1];
 
        // Check whether the current
        // sub-array yeilds maximum sum
        res = Math.Max(res, dp[i, m - 1]);
    }
 
    // Build the [,]dp matrix from
    // bottom to the top row
    for (i = n - 2; i >= 0; i--)
    {
        for (j = m - 2; j >= 0; j--)
        {
 
            // Update sum at each
            // cell in [,]dp
            dp[i, j] = mat[i, j] + dp[i, j + 1] +
                    dp[i + 1, j] - dp[i + 1, j + 1];
 
            // Update the maximum sum
            res = Math.Max(res, dp[i, j]);
        }
    }
 
    // Return the maximum sum
    return res;
}
 
// Driver Code
public static void Main(String[] args)
{
    // Given matrix [,]mat
    int [,]mat= {{ -6, -4, -1 },
                 { -3, 2, 4 },
                 { 2, 5, 8 } };
 
    // Function Call
    Console.Write(maxSubMatSum(mat));
}
}
 
// This code is contributed by Rohit_ranjan

Javascript

 
// Javascript program for the above approach   
 
 
// Function that finds the maximum
    // Sub-Matrix Sum
    function maxSubMatSum(mat)
    {
        // Number of rows in the matrix
        var n = mat.length;
 
        // Number of columns in the matrix
        var m = mat[0].length;
 
        var i, j;
 
        // dp matrix to store the
        // results of each iteration
        var dp = Array(n).fill().map(() =>
        Array(m).fill(0));
 
        // Base Case - The largest
        // element in the matrix
        dp[n - 1][m - 1] = mat[n - 1][m - 1];
 
        // To stores the final result
        var res = dp[n - 1][m - 1];
 
        // Find the max sub matrix sum for
        // the last row
        for (i = m - 2; i >= 0; i--) {
            dp[n - 1][i] = mat[n - 1][i] +
                            dp[n - 1][i + 1];
 
            // Check whether the current
            // sub-array yeilds maximum sum
            res = Math.max(res, dp[n - 1][i]);
        }
 
        // Calculate the max sub matrix
        // sum for the last column
        for (i = n - 2; i >= 0; i--) {
            dp[i][m - 1] = mat[i][m - 1] +
                            dp[i + 1][m - 1];
 
            // Check whether the current
            // sub-array yeilds maximum sum
            res = Math.max(res, dp[i][m - 1]);
        }
 
        // Build the dp matrix from
        // bottom to the top row
        for (i = n - 2; i >= 0; i--) {
            for (j = m - 2; j >= 0; j--) {
 
                // Update sum at each
                // cell in dp
                dp[i][j] = mat[i][j] + dp[i][j + 1] +
                dp[i + 1][j] - dp[i + 1][j + 1];
 
                // Update the maximum sum
                res = Math.max(res, dp[i][j]);
            }
        }
 
        // Return the maximum sum
        return res;
    }
 
    // Driver Code
     
        // Given matrix mat
        var mat = [ [ -6, -4, -1 ],
                    [ -3, 2, 4 ],
                    [ 2, 5, 8 ] ];
 
        // Function Call
        document.write(maxSubMatSum(mat));
 
// This code contributed by umadevi9616
 

输出：

19

时间复杂度： O(N*M)，其中 N 是行数，M 是列数
辅助空间： O(N*M)

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