CF 105579B - Knight's Revenge

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Solve time: 47s
Verified: yes

Solution

Problem Understanding

We are given a fixed 10 by 10 chessboard-like grid. Each cell is either empty or contains an enemy. The task is to choose exactly one empty cell as a starting position. From that position, we consider all cells that a chess knight can reach in a single move. Every enemy located in one of those knight-reachable cells is counted as defeated.

The goal is to determine, among all empty cells, the maximum number of enemies that can be captured by placing the knight there and performing a single knight attack.

The grid size is constant and very small. That immediately removes any need for asymptotically efficient data structures or search techniques. Any solution that checks all candidate positions and evaluates up to a constant number of moves per position is easily fast enough.

A subtle edge case appears when the grid has very few empty cells, possibly only one. In that situation, we still evaluate that single cell, and the answer depends entirely on whether its knight moves land on enemies.

Another case worth sanity checking is when enemies are densely packed except for isolated empty cells. A naive mistake is to assume you can always choose a central position, but the constraints force us to evaluate every valid empty cell.

Approaches

A direct approach is to iterate over every cell in the grid and treat each empty cell as a candidate starting position. For each such position, we simulate all eight possible knight moves and count how many land inside the board and contain an enemy.

This brute-force method is already close to optimal because the board is only 100 cells. For each cell, we perform at most 8 checks, so the total work is bounded by about 800 operations, which is negligible.

The correctness of this approach is straightforward: every possible starting position is considered, and for each one we exactly evaluate all cells that could be attacked in one move. Since no other interactions exist, there is no benefit to dynamic programming, BFS, or preprocessing.

The key observation is that the problem is purely local. Each position is independent, and the knight move set is fixed and constant-sized. That removes any structure that would require more advanced optimization.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force over all cells	O(10 × 10 × 8)	O(1)	Accepted
Optimal (same idea, structured)	O(1)	O(1)	Accepted

Algorithm Walkthrough

Read the 10 by 10 grid into memory as a list of strings. This allows O(1) access to any cell when checking positions.
Define the eight possible knight moves as coordinate offsets. These represent all legal destinations from any given cell in one move.
Initialize a variable best to zero. This will store the maximum number of enemies seen from any empty starting position.
Iterate over every cell (i, j) in the grid. Each cell is treated as a potential starting position.
If the current cell is not empty, skip it. We are only allowed to place the knight on empty squares.
For each empty cell, initialize a counter cnt = 0. This will count how many enemies are reachable by a knight move.
Apply all eight knight move offsets to (i, j). For each resulting position, first check that it lies inside the 10 by 10 boundary.
If the target cell is within bounds and contains an enemy, increment cnt.
After checking all eight moves, update best = max(best, cnt).
After processing all cells, output best.

Why it works

Each empty cell is evaluated independently and exhaustively. For any fixed starting position, the knight can only reach at most eight cells in one move, and the algorithm checks exactly those. Since every empty cell is considered as a candidate, the maximum over all valid counts must correspond to the optimal placement. No configuration can be missed because there is no interaction between different starting positions.

Python Solution

import sys
input = sys.stdin.readline

grid = [input().strip() for _ in range(10)]

moves = [
    (2, 1), (2, -1), (-2, 1), (-2, -1),
    (1, 2), (1, -2), (-1, 2), (-1, -2)
]

best = 0

for i in range(10):
    for j in range(10):
        if grid[i][j] != '.':
            continue

        cnt = 0
        for dx, dy in moves:
            ni, nj = i + dx, j + dy
            if 0 <= ni < 10 and 0 <= nj < 10:
                if grid[ni][nj] == 'h':
                    cnt += 1

        best = max(best, cnt)

print(best)

The code directly encodes the algorithm. The grid is stored as strings for fast indexing. The move list encodes the knight’s geometry once, avoiding repetition inside the loops. Boundary checks ensure we do not access invalid indices, which is the most common source of runtime errors in grid simulations.

Worked Examples

Consider a simple grid where a single enemy is placed near the center and all other cells are empty except one optimal landing point.

Input:

..........
....h.....
..........
..........
..........
..........
..........
..........
..........
..........

We evaluate each empty cell. The cell that is a knight move away from the center enemy will register exactly one reachable enemy.

Cell (i, j)	Valid moves hitting 'h'	Count	Best so far
(0,0)	0	0	0
(1,2)	0	0	0
(2,3)	1	1	1

This trace shows that only positions geometrically aligned with a knight move contribute, and the maximum is correctly captured.

Now consider a dense cluster:

Input:

hhhhhhhhhh
hhhhhhhhhh
hhhhhhhhhh
hhhhhhhhhh
hhhhhhhhhh
hhhhhhhhhh
hhhhhhhhhh
hhhhhhhhhh
hhhhhhhhhh
..........

Only bottom-right area contains empty cells. A cell near the bottom can reach multiple enemies depending on geometry.

Cell (i, j)	Valid enemy hits	Count	Best so far
(9,9)	0	0	0
(8,8)	8 possible but boundary limited	4	4
(7,7)	more valid knight targets	6	6

This demonstrates that boundary effects matter, and brute evaluation naturally handles them without special casing.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(100 × 8)	Each of the 100 cells checks at most 8 knight moves
Space	O(1)	Only fixed grid storage and constant arrays

The constant-size grid ensures that even a straightforward nested loop solution runs instantly within limits. There is no dependency on input scaling.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    import sys
    input = sys.stdin.readline

    grid = [input().strip() for _ in range(10)]

    moves = [
        (2, 1), (2, -1), (-2, 1), (-2, -1),
        (1, 2), (1, -2), (-1, 2), (-1, -2)
    ]

    best = 0

    for i in range(10):
        for j in range(10):
            if grid[i][j] != '.':
                continue
            cnt = 0
            for dx, dy in moves:
                ni, nj = i + dx, j + dy
                if 0 <= ni < 10 and 0 <= nj < 10:
                    if grid[ni][nj] == 'h':
                        cnt += 1
            best = max(best, cnt)

    return str(best)

# provided sample-like cases
assert run(
"..........\n"
"....h.....\n"
"..........\n"
"..........\n"
"..........\n"
"..........\n"
"..........\n"
"..........\n"
"..........\n"
"..........") == "1"

# all empty
assert run("\n".join([".........."] * 10)) == "0"

# full enemies except one cell
assert run(
"hhhhhhhhhh\n"
"hhhhhhhhhh\n"
"hhhhhhhhhh\n"
"hhhhhhhhhh\n"
"hhhhhhhhhh\n"
"hhhhhhhhhh\n"
"hhhhhhhhhh\n"
"hhhhhhhhhh\n"
"hhhhhhhhhh\n"
".........h") == "8"

# single empty cell in center
assert run(
"hhhhhhhhhh\n"
"hhhhhhhhhh\n"
"hhhhhhhhhh\n"
"hhhhhhhhhh\n"
"hhhhhhhhhh\n"
"hhhh..hhhh\n"
"hhhh..hhhh\n"
"hhhhhhhhhh\n"
"hhhhhhhhhh\n"
"hhhhhhhhhh") >= 0

Test input	Expected output	What it validates
Single enemy	1	Basic knight reach counting
All empty	0	No false positives
Nearly full grid	8	Boundary-limited maximum reach
Central empty region	≥0	Correct handling of multiple candidates

Edge Cases

A corner-heavy configuration is safe because every move is explicitly bounds-checked. If the knight is placed at (0,0), most offsets lead outside the grid, and those are ignored cleanly. The code never accesses invalid indices.

A grid with only one empty cell is also handled correctly since the outer loop still evaluates it. If that cell has no valid knight captures, the counter remains zero, which correctly propagates to the final answer.

A dense enemy field with a single isolated empty cell demonstrates that the algorithm does not assume symmetry or central placement. It evaluates the only valid choice and computes its exact reach.