思路：
状态压缩dp。
实现：
1 #include <iostream> 2 using namespace std; 3 const int MOD = 100000000; 4 int m, n, a[13][13], dp[13][13][1 << 13]; 5 int main() 6 { 7 cin >> m >> n; 8 for (int i = 0; i < m; i++) 9 { 10 for (int j = 0; j < n; j++) 11 { 12 cin >> a[i][j]; 13 } 14 } 15 for (int i = 0; i < 1 << n; i++) 16 { 17 bool flg = true; 18 for (int j = 0; j < n; j++) 19 { 20 if (a[m - 1][j] == 0 && i >> j & 1) 21 { 22 flg = false; 23 break; 24 } 25 } 26 dp[m][0][i] = flg; 27 } 28 for (int i = m - 1; i >= 0; i--) 29 { 30 for (int j = n - 1; j >= 0; j--) 31 { 32 for (int k = 0; k < 1 << n; k++) 33 { 34 int ni = i, nj = j + 1; 35 if (j == n - 1) { ni = i + 1; nj = 0; } 36 dp[i][j][k] += dp[ni][nj][k & ~(1 << j)]; 37 dp[i][j][k] %= MOD; 38 bool ok = true; 39 if (a[i][j] == 0) ok = false; 40 if (j && k >> (j - 1) & 1) ok = false; 41 if (k >> j & 1) ok = false; 42 if (ok) 43 { 44 dp[i][j][k] += dp[ni][nj][k | 1 << j]; 45 dp[i][j][k] %= MOD; 46 } 47 } 48 } 49 } 50 cout << dp[0][0][0] << endl; 51 return 0; 52 }