# Dynamic Programming on Broken Profile. Problem "Parquet"

Common problems solved using DP on broken profile include:

• finding number of ways to fully fill an area (e.g. chessboard/grid) with some figures (e.g. dominoes)
• finding a way to fill an area with minimum number of figures
• finding a partial fill with minimum number of unfilled space (or cells, in case of grid)
• finding a partial fill with the minimum number of figures, such that no more figures can be added

## 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: $dp[i, mask]$, where $i = 1, \ldots N$ and $mask = 0, \ldots 2^M - 1$.

$i$ 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 $dp[N, 0]$.

We will be building the DP state by iterating over each $i = 1, \cdots N$ and each $mask = 0, \ldots 2^M - 1$, 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> > dp;

void calc (int x = 0, int y = 0, int mask = 0, int next_mask = 0)
{
if (x == n)
return;
if (y >= m)
else
{
int my_mask = 1 << y;
else
{
}
}
}

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

dp.resize (n+1, vector<long long> (1<<m));
dp = 1;
for (int x=0; x<n; ++x)