(Analysis by Benjamin Qi)

Suppose that sweet corn grows in $(1,1)$. Consider the minimum $j$ such that alfalfa grows in $(1,j)$.


In general, an assignment of sweet corn or alfalfa to each square corresponds to a down-right path from $(1,1)$ to some square $(x,y)$ that satisfies $x=N+1$ or $y=N+1$. In the above example, the first three squares of the path are $(1,1)\to (1,j)\to (k,j)$. The squares that are just before where the path changes direction (such as $(1,j-1)$) must contain a sprinkler of a certain type (so their states are fixed), while every other square that does not contain a cow can be in one of two states: either place no sprinkler or place a sprinkler of the same type as the crop that grows in that square. A path that changes direction $d$ times fixes the states of $d+1$ squares, so the states of the remaining squares can be assigned in $2^{\text{(# unoccupied squares)}-d-1}$ ways. It suffices to sum $2^{-d-1}$ over all paths and then multiply the answer by $2^{\text{(# unoccupied squares)}}$ at the end. In the code below, $p\equiv 2^{-1}\pmod{10^9+7}$.

We can do this naively in $O(N^3)$ and use prefix sums to get $O(N^2)$. It is probably easier to write the $O(N^3)$ solution first and then figure out how to optimize it.

Dhruv Rohatgi's code:

#include <iostream>
#include <algorithm>
#include <cstdio>
using namespace std;
#define MOD 1000000007
int N;
long long p = 500000004LL;
char A[2005][2005];
char B[2005][2005];
int r[2005][2005];
int b[2005][2005];
int psr[2005][2005];
int psb[2005][2005];
int main()
	cin >> N;
	for(int i=0;i<N;i++)
		cin >> (A[i+1]+1);
	for(int i=2;i<=N+1;i++)
		if(A[i-1][1] == '.')
			b[i][0] = psb[i][0] = p;
	for(int j=1;j<=N;j++)
		if(A[1][j] == '.')
			r[1][j] = psr[1][j] = p;
	for(int i=2;i<=N+1;i++)
		for(int j=1;j<=N;j++)
			if(A[i][j] == '.')
				r[i][j] = (p*psb[i][j-1])%MOD;
			if(A[i-1][j+1] == '.')
				b[i][j] = (p*psr[i-1][j])%MOD;
			psr[i][j] = (psr[i-1][j] + r[i][j])%MOD;
			psb[i][j] = (psb[i][j-1] + b[i][j])%MOD;
	int ans = (psr[N][N] + psb[N+1][N])%MOD;
	for(int i=1;i<=N;i++)
		for(int j=1;j<=N;j++)
				ans = (2LL*ans)%MOD;
	cout << ans << '\n';