Bicolorings - codeforce
题解:dp[ i ][ k ][0 ~ 3] :表示第 i 列状态为s(s = 0,1,2,3)时总的联通个数。
注意:dp[ i ][ k ][ 1 ] 和 dp[ i ][ k ][ 2 ] 的转移。
const int N = 1005; const int mod = 998244353; int n, k; LL dp[N][2 * N][4]; int main() { cin >> n >> k; dp[1][1][0] = 1; dp[1][1][3] = 1; dp[1][2][1] = 1; dp[1][2][2] = 1; Rep(i, 2, n) Rep(j, 1, 2 * i) { dp[i][j][0] = dp[i - 1][j][0] + dp[i - 1][j][1] + dp[i - 1][j][2] + dp[i - 1][j - 1][3]; dp[i][j][3] = dp[i - 1][j - 1][0] + dp[i - 1][j][1] + dp[i - 1][j][2] + dp[i - 1][j][3]; if (j > 1) dp[i][j][1] = dp[i - 1][j - 1][0] + dp[i - 1][j][1] + dp[i - 1][j - 2][2] + dp[i - 1][j - 1][3]; if (j > 1) dp[i][j][2] = dp[i - 1][j - 1][0] + dp[i - 1][j - 2][1] + dp[i - 1][j][2] + dp[i - 1][j - 1][3]; rep(t, 0, 4) dp[i][j][t] %= mod; } LL res = (dp[n][k][0] + dp[n][k][1] + dp[n][k][2] + dp[n][k][3]) % mod; cout << res << endl; return 0; }