Rectangles

A valid rectangle only cares about its four corners. The inside can be a cursed wasteland of zeroes; nobody asked.

Pick two rows $u<d$. A column $c$ can be a vertical side of a rectangle with these rows iff both $G_{u,c}$ and $G_{d,c}$ are 1.

So for this row pair, collect all useful columns:

$c_1<c_2<\dots<c_k.$

Any two of them make a valid rectangle. But we do not need all pairs.

Key compression

For a fixed row pair, only adjacent useful columns matter:

$(u,d,c_i,c_{i+1}).$

Why? Take any rectangle using useful columns $c_a$ and $c_b$ with $a<b$. For any column $j$ inside it, there is some adjacent pair $c_i,c_{i+1}$ with

$c_i\le j\le c_{i+1}.$

That adjacent-pair rectangle has the same height and no larger width, so its area is no larger. Therefore the wider rectangle is never needed to get a minimum answer.

This gives a minimal/exhaustive set of rectangles. For every row pair, scan columns once and emit rectangles between consecutive common-1 columns.

Transpose trick

The expensive part is enumerating row pairs. If $n>m$, transpose the grid and solve the same problem. At the end, transpose the answer back.

Now the working grid has dimensions $R\times C$ where

$R=\min(n,m),\qquad C=\max(n,m).$

The number of row-pair/column checks is

$O(R^2C)=O(nm\min(n,m)).$

Since $nm\le 250000$, this is fine.

How to spread rectangle areas to cells

For each working column $j$, define:

$dp[u][d] =$ minimum area among generated rectangles with exact row span $[u,d]$ that cover column $j$.

When generating a rectangle $(u,d,l,r)$ with area $A$, it covers every column $j\in[l,r]$, so we do:

$dp_j[u][d]=\min(dp_j[u][d], A).$

Now, for a fixed column $j$, if a rectangle covers row interval $[u,d]$, it also covers every smaller interval inside it. In particular, it covers each single row $i$ with $u\le i\le d$.

So we propagate values from larger intervals to smaller intervals:

$dp[u][d] \leftarrow \min(dp[u][d], dp[u-1][d])$

if $u>0$, and

$dp[u][d] \leftarrow \min(dp[u][d], dp[u][d+1])$

if $d+1<R$.

Process intervals by decreasing length, so larger intervals are already finalized before smaller ones use them.

After propagation, $dp[i][i]$ is exactly the minimum area of any generated rectangle covering cell $(i,j)$. If it is still infinity, no rectangle covers that cell, so output 0.

Correctness

Every generated rectangle is valid because its two chosen rows both contain 1 in its two chosen columns.

Now take any valid rectangle $(u,d,l,r)$ covering some cell $(i,j)$. For the row pair $(u,d)$, both $l$ and $r$ are useful columns. Among the useful columns, choose the adjacent pair $a,b$ with $a\le j\le b$. Then $(u,d,a,b)$ is generated, covers $(i,j)$, and has width at most $r-l+1$. Its height is the same, so its area is no larger.

Thus generated rectangles are enough to achieve the optimum for every cell.

For a fixed column, initializing $dp[u][d]$ stores all generated rectangles with exact vertical span $[u,d]$. The decreasing-length propagation considers all larger intervals containing each smaller interval. Therefore $dp[i][i]$ becomes the minimum area among all generated rectangles whose row span contains $i$ and whose column span contains $j$.

By the previous paragraph, that is exactly the required answer.

Complexity

Let $R=\min(n,m)$ and $C=\max(n,m)$.

Enumeration and DP both take

$O(R^2C)=O(nm\min(n,m)).$

Memory is $O(R^2C)$ in this implementation, which fits the $1024$ MB limit under the given constraints.