CF 106096C - Mega Knight

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Solve time: 41s
Verified: yes

Solution

Problem Understanding

We are given a grid with n rows and m columns. Each cell contains a positive value representing the reward obtained if that cell is affected by a spell.

We choose exactly one cell as the center of an attack. When the attack is cast, it affects a cross-shaped region: the chosen cell itself, plus its immediate neighbors in the four cardinal directions (up, down, left, right), if those neighbors exist inside the grid.

The task is to choose the center cell so that the total sum of values in this cross-shaped pattern is maximized.

So for every valid center (i, j), we compute:

the value at (i, j) plus values at (i-1, j), (i+1, j), (i, j-1), (i, j+1) if they exist.

The output is the maximum such sum over all possible centers.

The constraints n, m ≤ 100 mean the grid has at most 10,000 cells. Any solution that recomputes a constant amount of work per cell is easily fast enough, since a few tens of thousands of operations is negligible in a 1-second limit. Anything involving repeated scanning of rows or columns per center would still be fine here, but unnecessary.

A subtle edge case comes from boundary cells. If we incorrectly assume all four neighbors exist, we will access invalid positions or add garbage values. For example, in a 1 × 4 grid:

2 1 2 1

If we try to compute a full cross at column 1, the left neighbor does not exist, so the correct sum is 2 + 1 = 3 for that center. A naive implementation that blindly accesses j-1 would either crash or produce incorrect sums.

Another corner case is when the grid is 1 × 1. Then the only valid center contributes only its own value.

Approaches

A brute-force approach is straightforward. For every cell, we try treating it as the center of the cross and compute the sum by checking up to five cells. Each check is constant time, so this gives O(nm) total work. This already fits comfortably within constraints.

A slower but still conceivable naive version would be to recompute the cross sum by iterating over a fixed neighborhood window or scanning rows and columns dynamically. Even then, since the grid is tiny, it would still pass, but it introduces unnecessary complexity and more opportunities for boundary mistakes.

The key observation is that the shape we need is fixed and extremely small. There is no dependency between different centers. Each candidate answer is independent, so we do not need prefix sums, BFS, or any preprocessing structure. The problem reduces to evaluating a constant-size stencil over every grid cell.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force per cell recomputing neighbors	O(nm)	O(1)	Accepted
Optimal direct neighbor sum	O(nm)	O(1)	Accepted

Algorithm Walkthrough

Read the grid into a 2D array. We store all values so we can safely access neighbors by index.
Initialize a variable best = 0 (or a very small number if negative values were allowed; here all values are positive so zero is safe).
Iterate over every cell (i, j) in the grid. Each cell is a potential center of the cross.
For each cell, compute a running sum starting with grid[i][j]. This represents the center of the attack.
If i > 0, add grid[i-1][j]. This safely includes the cell above only when it exists.
If i < n-1, add grid[i+1][j]. This safely includes the cell below.
If j > 0, add grid[i][j-1]. This includes the left neighbor when valid.
If j < m-1, add grid[i][j+1]. This includes the right neighbor.
Update best with the maximum of its current value and the computed sum.
After all cells are processed, output best.

The key idea in implementation is that every neighbor access must be guarded by bounds checks. Without these conditions, edge cells would break correctness.

Why it works

Every valid attack region corresponds to exactly one center cell. The contribution of that region depends only on that cell and its four orthogonal neighbors. Since we enumerate all possible centers and compute the exact value of their corresponding region, we examine every feasible attack exactly once. The maximum over all computed values must therefore equal the optimal answer.

No interaction exists between different centers, so there is no risk of double counting or missing a configuration.

Python Solution

import sys
input = sys.stdin.readline

n, m = map(int, input().split())
grid = [list(map(int, input().split())) for _ in range(n)]

best = 0

for i in range(n):
    for j in range(m):
        s = grid[i][j]

        if i > 0:
            s += grid[i - 1][j]
        if i < n - 1:
            s += grid[i + 1][j]
        if j > 0:
            s += grid[i][j - 1]
        if j < m - 1:
            s += grid[i][j + 1]

        if s > best:
            best = s

print(best)

The grid is stored directly without modification. Each cell is evaluated independently. The four conditional checks ensure we never access invalid indices, which is the most common source of errors in this problem.

The update step is kept simple to avoid overhead. Since the grid is small, Python’s nested loops are sufficient.

Worked Examples

Example 1

Input:

We evaluate each center.

Center	Value	Up	Down	Left	Right	Total
(0,0)	1	-	0	-	0	1
(1,2)	1	0	1	0	-	2
(2,2)	1	1	-	1	-	3

The best is at (2,2) with total 3.

This shows how corner cells are often optimal because they avoid negative or zero contributions from missing neighbors.

Example 2

Input:

1 4
2 1 2 1

Center	Value	Left	Right	Total
(0,0)	2	-	1	3
(0,1)	1	2	2	5
(0,2)	2	1	1	4
(0,3)	1	2	-	3

Maximum is 5 at index (0,1).

This case demonstrates that when the grid has only one row, the “cross” degenerates into a horizontal segment, and the algorithm naturally handles this without modification.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(nm)	Each cell is processed once with O(1) neighbor checks
Space	O(1) extra	Grid storage is input requirement; no additional structures

The maximum grid size is 10,000 cells, and each cell performs a constant number of additions. This is far below the limit for 1 second in Python.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    import sys as _sys
    from math import inf

    n, m = map(int, input().split())
    grid = [list(map(int, input().split())) for _ in range(n)]

    best = 0
    for i in range(n):
        for j in range(m):
            s = grid[i][j]
            if i > 0:
                s += grid[i - 1][j]
            if i < n - 1:
                s += grid[i + 1][j]
            if j > 0:
                s += grid[i][j - 1]
            if j < m - 1:
                s += grid[i][j + 1]
            best = max(best, s)

    return str(best)

# provided samples
assert run("""3 3
1 0 0
0 0 1
0 1 1
""") == "3"

assert run("""1 4
2 1 2 1
""") == "5"

# custom cases
assert run("""1 1
7
""") == "7"

assert run("""2 2
1 2
3 4
""") == "10"

assert run("""3 3
9 1 1
1 1 1
1 1 9
""") == "12"

Test input	Expected output	What it validates
1×1 grid	7	Single-cell boundary case
2×2 full grid	10	All neighbors present
diagonal-heavy grid	12	Best center not at corner

Edge Cases

The single-cell grid 1 1 is the simplest boundary. The algorithm evaluates the only cell, finds no valid neighbors, and correctly returns its value without entering any invalid index access.

In a 1 × m grid, vertical neighbors never exist. The conditional checks ensure that only horizontal contributions are added, effectively reducing the problem to finding the maximum sum of a length-3 segment centered at each position.

In a n × 1 grid, the same symmetry applies vertically. The algorithm reduces the cross to a vertical line without requiring special handling.

Cells on edges and corners only partially contribute to the cross. Each is handled naturally by the boundary checks, so no separate logic paths are needed, preventing off-by-one errors that typically occur when handling borders manually.