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To solve this problem, we need to maximize the sum of the elements in an n x m grid after performing any number of row and column flips. The key insight is to use row flips to make the first element of each row positive, then use column flips to maximize the sum of each subsequent column.

Approach

1. Row Flips: For each row, flip the row if the first element is negative. This ensures the first element of each row is positive, maximizing the sum of the first column (since all elements are positive).
2. Column Flips: For each column starting from the second column, if the sum of the elements in the column is negative, flip the column to make the sum positive (since flipping a column with a negative sum will turn it into a positive sum, which increases the total sum). Instead of explicitly flipping the column, we can simply add the absolute value of the column sum to the total sum (as flipping the column is equivalent to taking the absolute value of the sum).

Solution Code

n, m = map(int, input().split())
grid = []
for _ in range(n):
    row = list(map(int, input().split()))
    if row[0] < 0:
        row = [-x for x in row]
    grid.append(row)

total_sum = sum(row[0] for row in grid)

for j in range(1, m):
    col_sum = sum(row[j] for row in grid)
    total_sum += abs(col_sum)

print(total_sum)

Explanation

1. Row Flips: By flipping rows where the first element is negative, we ensure the first column's elements are all positive, so their sum is the maximum possible (sum of absolute values of the original first elements).
2. Column Flips: For each subsequent column, the sum of elements can be maximized by taking the absolute value of the sum (since flipping the column if the sum is negative will give a positive sum). This step ensures each column contributes the maximum possible value to the total sum.

This approach efficiently computes the maximum sum with a time complexity of O(n*m), which is optimal for the problem constraints. The solution is both correct and easy to implement.

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