The Turtle Strikes Back

For a fixed cell $x=(i,j)$, after Raphael flips it, Michelangelo's best path is the maximum of two kinds of paths:

1. paths that use $x$;
2. paths that avoid $x$.

The first kind is standard. Let

$pref[i][j]=\max\text{ sum from }(1,1)\text{ to }(i,j),$

and

$suf[i][j]=\max\text{ sum from }(i,j)\text{ to }(n,m).$

Then the best original path forced to pass through $(i,j)$ is

$through[i][j]=pref[i][j]+suf[i][j]-a_{i,j}.$

If Raphael flips this cell, every such path loses $2a_{i,j}$, so the best path using it becomes

$through[i][j]-2a_{i,j}.$

Now we need the best path that avoids $(i,j)$.

A monotone path from top-left to bottom-right visits exactly one cell on every anti-diagonal $i+j=s$. So if a path avoids $(i,j)$, then on anti-diagonal $s=i+j$ it must pass through some other cell. Conversely, if it passes through another cell on that same anti-diagonal, it cannot also pass through $(i,j)$.

So the best path avoiding $(i,j)$ is just the best complete path through any other cell on the same anti-diagonal:

$avoid[i][j]=\max_{(r,c):\ r+c=i+j,\ (r,c)\ne(i,j)} through[r][c].$

For each anti-diagonal, compute all $through$ values on it. To get “best except this cell” quickly, use prefix and suffix maxima along that diagonal. For position $k$ inside the diagonal list:

$avoid=\max(prefix[k-1], suffix[k+1]).$

If there is no other cell on that diagonal, then avoiding this cell is impossible, so $avoid=-\infty$. This matters for a $1\times1$ grid: Michelangelo must eat the only slice.

For a candidate sauced cell $(i,j)$, the resulting best pleasure is

$candidate(i,j)=\max\left(avoid[i][j],\ through[i][j]-2a_{i,j}\right).$

Raphael wants the minimum candidate.

We only need to consider cells with $a_{i,j}\ge 0$. If $a_{i,j}<0$, flipping it makes that cell larger, so every path becomes no smaller than before. Since the statement guarantees at least one non-negative cell, choosing a negative cell is never optimal.

Correctness proof:

First, for any cell $x$, $through[x]$ is the maximum original sum of a full path that passes through $x$. Any such path is a best prefix ending at $x$ plus a best suffix starting at $x$, with $a_x$ counted twice, so subtracting it once gives exactly the best full path through $x$.

After flipping $x$, every path through $x$ has its sum changed by $-2a_x$. Therefore the best flipped path among paths through $x$ is $through[x]-2a_x$.

Next, every valid path visits exactly one cell on each anti-diagonal $i+j=s$, because each move increases $i+j$ by exactly $1$. A path avoids $x$ if and only if, on $x$'s anti-diagonal, it uses a different cell. For any other cell $y$ on that anti-diagonal, the best path through $y$ has value $through[y]$ and certainly avoids $x$. Thus the best path avoiding $x$ is the maximum $through[y]$ over all other cells $y$ on the same anti-diagonal.

So after flipping $x$, Michelangelo's optimal value is exactly

$\max(avoid[x], through[x]-2a_x).$

The algorithm evaluates this exact expression for every relevant cell, so it returns the value Raphael can guarantee.

Complexity: both DP passes are $O(nm)$, and every cell is processed once more inside its anti-diagonal. Memory is $O(nm)$ and total time is $O(nm)$ per test case.