裸的矩阵树定理。求行列式的时候答案要在中间统计,因为交换两个行会使答案取反。
#include <bits/stdc++.h>
using namespace std;
const int N = 10 + 5;
const int M = 100 + 5;
const int mod = 1000000000;
char s[N][N];
int n, m, a[M][M], id[N][N];
void add (int u, int v) {
--a[u][v], --a[v][u], ++a[u][u], ++a[v][v];
}
int Gauss (int n) {
int ans = 1;
for (int i = 1; i <= n; ++i) {
for (int k = i + 1; k <= n; ++k) {
while (a[k][i]) {
int d = a[i][i] / a[k][i];
for (int j = i; j <= n; ++j) {
a[i][j] = (((a[i][j] - 1LL * d * a[k][j]) %mod) + mod) %mod;
}
swap (a[i], a[k]), ans =- ans;
}
}
ans = (((1LL * ans * a[i][i]) % mod) + mod) % mod;
}
return ans;
}
int main () {
cin >> n >> m;
for (int i = 1; i <= n; ++i) cin >> s[i] + 1;
int idx = 0;
for (int i = 1; i <= n; ++i) {
for (int j = 1; j <= m; ++j) {
if (s[i][j] == '.') {
id[i][j] = ++idx;
}
}
}
for (int i = 1; i <= n; ++i) {
for (int j = 1; j <= m; ++j) {
if (s[i][j] == '.') {
if(id[i - 1][j]) add (id[i][j], id[i - 1][j]);
if(id[i][j - 1]) add (id[i][j], id[i][j - 1]);
}
}
}
cout << Gauss (idx - 1) << endl;
return 0;
}