Dynamic Programming on Broken Profile. Problem "Parquet"

Common problems solved using DP on broken profile include:

Problem "Parquet"

Problem description. Given a grid of size $N \times M$. Find number of ways to fill the grid with figures of size $2 \times 1$ (no cell should be left unfilled, and figures should not overlap each other).

Let the DP state be: D[i][mask], where i = 1 ... N, and mask = 0 ... 2^M-1. i respresents number of rows in the current grid, and mask is the state of last row of current grid. If j-th bit of mask is 0 then the corresponding cell is filled, otherwise it is unfilled. Clearly, the answer to the problem will be D[N][0].

We will be building the DP state by iterating over each i = 1 ... N and each mask = 0 ... 2^M-1, and for each mask we will be only transitioning forward, that is, we will be adding figures to the current grid.

Implementation

int n, m;
vector < vector<long long> > d;


void calc (int x = 0, int y = 0, int mask = 0, int next_mask = 0)
{
    if (x == n)
        return;
    if (y >= m)
        d[x+1][next_mask] += d[x][mask];
    else
    {
        int my_mask = 1 << y;
        if (mask & my_mask)
            calc (x, y+1, mask, next_mask);
        else
        {
            calc (x, y+1, mask, next_mask | my_mask);
            if (y+1 < m && ! (mask & my_mask) && ! (mask & (my_mask << 1)))
                calc (x, y+2, mask, next_mask);
        }
    }
}


int main()
{
    cin >> n >> m;

    d.resize (n+1, vector<long long> (1<<m));
    d[0][0] = 1;
    for (int x=0; x<n; ++x)
        for (int mask=0; mask<(1<<m); ++mask)
            calc (x, 0, mask, 0);

    cout << d[n][0];

}