CF 445A - DZY Loves Chessboard

CF 445A - DZY Loves Chessboard

Rating: 1200
Tags: dfs and similar, implementation
Solve time: 1m 21s
Verified: yes

Solution

Problem Understanding

We are given a chessboard represented as an n by m grid where each cell is either good or bad. A good cell is indicated by a "." and a bad cell by a "-". On every good cell, we must place a chessman that is either white ("W") or black ("B") such that no two chessmen of the same color share a side. Bad cells remain empty and are not considered in adjacency checks. The task is to produce any valid arrangement of the chessmen.

The input size is modest: both n and m are at most 100. This immediately rules out solutions with time complexity worse than O(n * m), since n * m can reach 10,000 and any algorithm with a quadratic factor or repeated deep recursion might be inefficient. Memory is also sufficient for a straightforward 2D array representation of the board.

A subtle edge case arises when a good cell is surrounded entirely by bad cells. For example, if the input is:

3 3
---
-.-
---

then the middle cell can be either "W" or "B". A naive adjacency-based approach that assumes all neighbors are good could misplace colors. Another edge case is a single-cell board, 1 1, where either color is valid. A careless checker might require neighbors to exist and fail on this minimal input.

Approaches

A brute-force approach would attempt to place a color on each good cell and then recursively or iteratively check all possible combinations while respecting adjacency constraints. This approach is correct but combinatorially explosive. For an n * m board with about half good cells, the number of placements is roughly 2^(n*m/2). Clearly, this exceeds any feasible runtime, even for n = m = 10, and is impossible for the upper bound of 100.

The key insight is that the chessboard is a bipartite structure: we can color the board in a checkerboard pattern such that no two adjacent cells share the same parity of row+column sum. Specifically, for a cell at position (i, j), if (i + j) is even we can place a "B", and if (i + j) is odd we place a "W". This pattern automatically guarantees that adjacent good cells have opposite colors. Bad cells are ignored, so they remain as "-" and do not affect the coloring.

This reduces the problem to a simple iteration over the grid, computing parity, and assigning the appropriate character based on the cell type and parity. No recursion or backtracking is necessary.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(2^(n*m))	O(n*m)	Too slow
Optimal	O(n*m)	O(n*m)	Accepted

Algorithm Walkthrough

1. Read the dimensions n and m from input and then the n strings representing the board. Each string has length m.
2. Initialize an empty output board as a list of strings or characters.
3. Iterate over each row i from 0 to n-1. For each row, iterate over each column j from 0 to m-1.
4. For each cell at (i, j), check if it is a bad cell ("-"). If so, copy it directly to the output.
5. If the cell is good (".") compute the sum i + j. If (i + j) % 2 == 0, place "B" on that cell; otherwise, place "W". The choice of which parity gets which color is arbitrary; the key is consistency across the board.
6. After processing all columns in a row, append the row to the output board.
7. Print the output board row by row.

Why it works: the parity of row + column ensures that no two adjacent cells share the same parity. Since we assign different colors to even and odd sums, any two neighboring good cells automatically have opposite colors. Bad cells are never altered, so adjacency rules are never violated by them. This guarantees a valid coloring for any board.

Python Solution

import sys
input = sys.stdin.readline

n, m = map(int, input().split())
board = [input().strip() for _ in range(n)]
output = []

for i in range(n):
    row = []
    for j in range(m):
        if board[i][j] == '-':
            row.append('-')
        elif (i + j) % 2 == 0:
            row.append('B')
        else:
            row.append('W')
    output.append(''.join(row))

print('\n'.join(output))

The code follows the algorithm directly. Reading the board uses sys.stdin.readline for fast input. Each row is built incrementally using a list of characters for efficiency, then joined into a string. The parity check (i + j) % 2 assigns the color consistently. Edge conditions such as single-cell boards, all-bad rows, or irregular patterns are handled naturally because we never assume neighbors exist beyond calculating (i + j).

Worked Examples

Sample Input 1

1 1
.

i	j	board[i][j]	i+j	parity	placed
0	0	.	0	even	B

Output:

B

This demonstrates minimal input handling. The single cell gets color based on parity.

Sample Input 2

2 2
..
..

i	j	board[i][j]	i+j	parity	placed
0	0	.	0	even	B
0	1	.	1	odd	W
1	0	.	1	odd	W
1	1	.	2	even	B

Output:

BW
WB

This confirms that the checkerboard pattern ensures adjacent cells differ in color.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n*m)	We visit each cell exactly once to compute parity and assign a color.
Space	O(n*m)	We store the output board, which has the same size as the input board.

Given n and m up to 100, the maximum of 10,000 operations is easily handled within 1 second and memory usage is well within the 256 MB limit.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    n, m = map(int, input().split())
    board = [input().strip() for _ in range(n)]
    output = []
    for i in range(n):
        row = []
        for j in range(m):
            if board[i][j] == '-':
                row.append('-')
            elif (i + j) % 2 == 0:
                row.append('B')
            else:
                row.append('W')
        output.append(''.join(row))
    return '\n'.join(output)

# Provided samples
assert run("1 1\n.") == "B"
assert run("2 2\n..\n..") == "BW\nWB"

# Custom cases
assert run("3 3\n-..\n.-.\n..-") == "-BW\nW-B\nBW-", "mixed bad cells"
assert run("1 5\n.....") == "BWBWB", "single row"
assert run("5 1\n.\n.\n.\n.\n.") == "B\nW\nB\nW\nB", "single column"
assert run("2 3\n---\n...") == "---\nBWB", "top bad row"
assert run("3 3\n...\n...\n...") == "BWB\nWBW\nBWB", "full 3x3"

Test input	Expected output	What it validates
3x3 mixed bad cells	-BW\nW-B\nBW-	Correct handling of bad cells interleaved with good cells
1x5 row	BWBWB	Correct single-row coloring
5x1 column	B\nW\nB\nW\nB	Correct single-column coloring
2x3 top bad row	---\nBWB	Ensures bad rows are preserved
3x3 full	BWB\nWBW\nBWB	Standard checkerboard coloring correctness

Edge Cases

A single-cell board:

1 1
.

The algorithm computes (0+0) % 2 = 0 and places "B". No adjacency exists, so the color is valid.

A cell surrounded entirely by bad cells:

3 3
---
-.-
---

Only cell (1,1) is good. (1+1)%2=2%2=0, so "B" is placed. No adjacent good cells exist, so the adjacency invariant trivially holds.

A full row or column of bad cells:

2 3
---
.W.

Bad cells remain "-", good cells are assigned based on parity. The algorithm places "W" at (1,1) because (1+1)%2=0, consistent with the parity rule. The adjacent bad cells do not interfere.