### ismail's blog

By ismail, history, 12 months ago,

## Problem Details:

Contest : Codeforces Round #589 (Div. 2)

Problem : 1228B - Заполнение таблицы

Solution : Submission

## Solution Explanation:

Create a matrix a[r+2][c+2] to hold cell information (2 extra index for 1 based indexing + no need to check overflow)

#### Cell information:

-1 = Free (can be White or Black)

0 = Reserved (Must be empty and immutable)

1 = Occupied (Black cell)

[ To make r and c valid, reserved cell never be changed ]

#### Logic:

1. Initialize all cell as free (-1). memset(a, -1, sizeof(a[0][0]) * (r + 2) * (c + 2));

2. Take input for each row (1 to h) as "d" and do the following:

• Make cell d+1 reserved :=> If d==0, then 1st cell (for 1 base indexing, cell d+1=1 is the 1st cell) must be reserved (0) otherwise cell d+1 reserved because if d+1 cell changed to black then r will be invalid.
• Then make all (1 to d) cell occupied (1)
3. Take input for each col (1 to w) as "d" and do the following:(Same as logic 2, just extra need to checkout reserved cell is occupied or not before making new cell reserved or occupied)

• Check required reserved cell is occupied or not. If occupied then it's impossible to make a grid to satisfy such r, c values, so return 0 with "0" output.
• Make cell d+1 reserved.
• Then make all (1 to d) cell occupied (1) BUT before made occupied check it previously reserved or not. If reserved then return 0 with "0" output.
4. Count how many free (-1) cell do you have for combinations. If free cell is zero then ans = 1(initialized) otherwise ans = 2^free_cell.

.

Solution / Code:

Happy coding

• +21

 » 12 months ago, # |   +3 Can you make a similar for the rest
 » 2 months ago, # |   0 Thanks for the editorial it has been of great help understanding the problem.