Greedy Grid Counting

The trick is that the grid is mostly a distraction. The useful object is not the full $2\times n$ grid, but the diagonal difference sequence $d_i=a_{1,i+1}-a_{2,i}\qquad 1\le i<n.$ That sequence completely decides whether the greedy path can be optimal in every $2$-row subgrid.

Reduce A Subgrid To One Sequence

Consider a subgrid using both rows and columns $[l,r]$. A down/right path starts at $(1,l)$, moves right on the top row, moves down once, then moves right on the bottom row.

So every path is determined by the column $t$ where it moves down. Its value is $P_t=\sum_{j=l}^{t}a_{1,j}+\sum_{j=t}^{r}a_{2,j}.$ Compare two neighboring choices: $P_{t+1}-P_t=a_{1,t+1}-a_{2,t}=d_t.$ So if $d_t>0$, moving the down step one column right improves the path. If $d_t<0$, it makes the path worse. If $d_t=0$, both are tied.

Now look at the greedy rule. At top cell $(1,t)$, the two possible next cells are:

• right: $a_{1,t+1}$,
• down: $a_{2,t}$.

So greedy can move right iff $d_t\ge 0$, and can move down iff $d_t\le 0$.

Same comparison. Not similar. Same. That's the whole problem cracking open.

What Makes One Interval Valid?

Inside interval $[l,r]$, suppose all differences $d_l,d_{l+1},\dots,d_{r-1}$ are nonnegative. Then $P_t$ is nondecreasing, so the best path moves down at $r$. Greedy can move right the whole way, so this interval is fine.

Otherwise, let $p$ be the first index in this interval with $d_p<0$.

Before $p$, all differences are nonnegative, so $P_l\le P_{l+1}\le\cdots\le P_p.$ Greedy can reach column $p$, and since $d_p<0$, it can move down there. Thus the greedy candidate has value $P_p$.

This greedy path is optimal exactly when no later path beats it: $P_q\le P_p\qquad \text{for all }q>p.$ Using the difference sequence, that becomes $d_p+d_{p+1}+\cdots+d_{q-1}\le 0\qquad \text{for all }q>p.$ So every prefix sum starting at the first negative must stay nonpositive.

Because the interval is arbitrary, we can start the interval exactly at any negative position. Therefore the full condition is:

For every $p$ with $d_p<0$, $\sum_{i=p}^{q}d_i\le 0\qquad \text{for every }q\ge p.$

That is both necessary and sufficient.

Turn The Condition Into A Small Automaton

Scan the differences from left to right.

Before we have seen any negative difference, everything is easy: all processed differences are nonnegative, and there is no restriction yet. Call this the free state.

Once we take a negative difference, say from current prefix height $S$ using $d=-g$, the future prefix height is never allowed to exceed the old height $S$. Instead of storing absolute heights, store the slack $h=\text{cap}-\text{current height}.$ After taking $d=-g$, the current height drops by $g$, so the slack becomes $g$.

From a bounded state with slack $h$:

• If the next difference is $d\ge 0$, it is allowed only if $d\le h$, and the new slack is $h-d$.
• If the next difference is $d=-g<0$, it is always allowed, and it resets the slack to $g$.

That reset is the nice part. We do not need prefix sums, caps, or any large coordinate compression circus. Just $h\in[0,k-1]$.

Weights For Each Difference

Difference $d_i$ uses only two cells: $a_{1,i+1}\quad\text{and}\quad a_{2,i}.$ These pairs are disjoint for different $i$, which is extremely convenient.

For each position $i$, compute $w_i(d)=\#\{(x,y):x-y=d\},$ where $x$ must match cell $a_{1,i+1}$ if fixed, and $y$ must match cell $a_{2,i}$ if fixed.

If both are free, then $w_i(d)=k-|d|$ for $-(k-1)\le d\le k-1$.

The cells $a_{1,1}$ and $a_{2,n}$ never appear in any $d_i$, so they are independent. Multiply the final answer by $k$ for each of those cells that is empty.

DP

Let:

• free = number of ways where no negative difference has appeared yet,
• dp[h] = number of ways in bounded mode with current slack $h$.

For each position $i$:

1. A nonnegative difference keeps free free: $free' = free\cdot \sum_{d=0}^{k-1} w_i(d).$

2. A negative difference $d=-g$ can be taken from any state and resets slack to $g$: $dp'[g] += (free+\sum_h dp[h])\cdot w_i(-g).$

3. From bounded slack $h$, a nonnegative $d$ is allowed only when $d\le h$: $dp'[h-d]+=dp[h]\cdot w_i(d).$

The answer after all differences is $free+\sum_h dp[h],$ then multiplied by the independent choices for $a_{1,1}$ and $a_{2,n}$.

Complexity is $O(nk^2)$ per test case, and since $\sum n\le 500$ and $k\le 500$, this is completely fine.