A problem that is simple to solve in one dimension is often much more difficult to solve in more than one dimension. Consider satisfying a boolean expression in conjunctive normal form in which each conjunct consists of exactly 3 disjuncts. This problem (3-SAT) is NP-complete. The problem 2-SAT is solved quite efficiently, however. In contrast, some problems belong to the same complexity class regardless of the dimensionality of the problem.

Given a 2-dimensional array of positive and negative integers, find the sub-rectangle with the largest sum. The sum of a rectangle is the sum of all the elements in that rectangle. In this problem the sub-rectangle with the largest sum is referred to as the *maximal sub-rectangle*. A sub-rectangle is any contiguous sub-array of size or greater located within the whole array. As an example, the maximal sub-rectangle of the array:

is in the lower-left-hand corner:

and has the sum of 15.

The input consists of an array of integers. The input begins with a single positive integer *N* on a line by itself indicating the size of the square two dimensional array. This is followed by integers separated by white-space (newlines and spaces). These integers make up the array in row-major order (i.e., all numbers on the first row, left-to-right, then all numbers on the second row, left-to-right, etc.). *N* may be as large as 100. The numbers in the array will be in the range [-127, 127].

The output is the sum of the maximal sub-rectangle.

**Sample Input**

4

0 -2 -7 0 9 2 -6 2

-4 1 -4 1 -1

8 0 -2

**Sample Output**

15

using System;

using System.Collections.Generic;

using System.Linq;

using System.Text;

namespace MaxSum

{

class Program

{

static void Main(string[] args)

{

//Array initialization

int[,] matrix = new int[,] {

{0, -2, -7, 0},

{9, 2, -6, 2},

{-4, 1, -4, 1},

{-1, 8, 0, -2}

};

int rows = matrix.GetLength(0);

int cols = matrix.GetLength(1);

int flag = 0, sum = 0, row = 0, col = 0, rowSize = 0, colSize = 0;

for(int i = 0; i < rows; i++)

{

for(int j = 0; j < cols; j++)

{

//i and j will determine the varying window size here

for (int p = 0; p < rows; p++)

{

for (int q = 0; q < cols; q++)

{

//p and q are the matrix elements

if (p + i < rows && q + j < cols)

{

//Call the sum function here

if (flag == 0)

{

sum = Sum(matrix, p, q, i, j);

rowSize = i;

colSize = j;

row = p;

col = q;

flag = 1;

}

else

{

if (sum < Sum(matrix, p, q, i, j))

{

sum = Sum(matrix, p, q, i, j);

rowSize = i;

colSize = j;

row = p;

col = q;

}

}

}

}

}

}

}

Console.WriteLine("Max Sum = " + sum);

Console.WriteLine("Maximal Sub Matrix is:");

for (int i = row; i <= row + rowSize; i++)

{

for (int j = col; j <= col + colSize; j++)

{

Console.Write(matrix[i, j]);

Console.Write(" ");

}

Console.WriteLine();

}

}

public static int Sum(int[,] matrix, int row, int col, int rowSize, int colSize)

{

int sum = 0;

for (int i = row; i <= row + rowSize; i++)

{

for (int j = col; j <= col + colSize; j++)

{

sum = sum + matrix[i, j];

}

}

return sum;

}

}

}

Max Sum = 15

Maximal Sub Matrix is:

9 2

-4 1

-1 8

Press any key to continue . . .