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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 $mask = 0, \ldots 2^M - 1$.

$i$ represents 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)  dp[x+1][next_mask] += dp[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;  dp.resize (n+1, vector<long long> (1<<m));  dp[0][0] = 1;  for (int x=0; x<n; ++x)  for (int mask=0; mask<(1<<m); ++mask)  calc (x, 0, mask, 0);  cout << dp[n][0]; }

Practice Problems

References