思路：

LGV 定理 (Lindström–Gessel–Viennot lemma)

从{

\(a_1\),\(a_2\),...,\(a_n\)} 到 {

\(b_1\),\(b_2\),...,\(b_n\)}的不相交路径数等于行列式

\[ {\left[ \begin{array}{ccc} c(a_1, b_1) & c(a_1, b_2) & ... & c(a_1, b_n) \\ c(a_2, b_1) & c(a_2, b_2) & ... & c(a_2, b_n) \\ ... & ... & ... & ... \\ c(a_n, b_1) & c(a_n, b_2) & ... &c(a_n, b_n) \\ \end{array} \right ]} \]

的值。其中，\(c(a_i, b_i)\) 表示从点 \(a_i\) 到点 \(b_i\) 的路径方案数。

那么这道题就是求一个二阶行列式的值

代码：

#pragma GCC optimize(2)#pragma GCC optimize(3)#pragma GCC optimize(4)#include
    
     using namespace std;#define y1 y11#define fi first#define se second#define pi acos(-1.0)#define LL long long//#define mp make_pair#define pb push_back#define ls rt<<1, l, m#define rs rt<<1|1, m+1, r#define ULL unsigned LL#define pll pair
     
      #define pli pair
      
       #define pii pair
       
        #define piii pair
        
         #define pdd pair
         
          #define mem(a, b) memset(a, b, sizeof(a))#define debug(x) cerr << #x << " = " << x << "\n";#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);//head const int N = 3e3 + 5;const int MOD = 1e9 + 7;int dp[N][N], n, m;char s[N][N];int solve(int a, int b, int c, int d) { for (int i = 1; i <= n; ++i) for (int j = 1; j <= m; ++j) dp[i][j] = 0; for (int i = a; i <= c; ++i) { for (int j = b; j <= d; ++j) { if(i == a && j == b) { if(s[i][j] == '.') dp[i][j] = 1; } else { if(s[i][j] == '.') dp[i][j] = (dp[i-1][j]+dp[i][j-1])%MOD; } } } return dp[c][d];}int main() { scanf("%d %d", &n, &m); for (int i = 1; i <= n; ++i) scanf("%s", s[i]+1); printf("%lld\n", (solve(1, 2, n-1, m)*1LL*solve(2, 1, n, m-1) - solve(1, 2, n, m-1)*1LL*solve(2, 1, n-1, m)%MOD+MOD)%MOD); return 0;}