Additional DP Optimizations and Techniques

Author: Andi Qu

Techniques and optimizations like Knuth's optimization.

Knuth's Optimization

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Miscellaneous Techniques


Miscellaneous techniques

"Connected Component" DP Problems

StatusSourceProblem NameDifficultyTags

DP on Broken Profile

Broken profile DP is a subset of bitmask DP. Problems falling under this category generally have the following properties:

  1. They're about filling a 2D grid.
  2. One of the dimensions is much smaller than the other.
  3. When filling the grid, each cell depends only on adjacent cells.
  4. The cells don't have many possible values (usually only 2).

The third property is especially important, as it means that we can process the cells column-by-column (imagine a snake wrapping around the grid). We then only need to care about the rightmost processed cell in each row (hence the name "broken profile").

The fourth property suggests that we should use a bitmask to represent that broken profile.


Note that the bitmask doesn't necessarily have to represent the state of the cells. In some problems (e.g. CEOI 2006 Connect), it can represent the state of the cell borders instead.

Focus Problem – read through this problem before continuing!


Time Complexity: O(NM2N)\mathcal O(NM 2^N).

We'll process the grid cells column-by-column, row-by-row.

Let dp[i][j][mask]\texttt{dp}[i][j][mask] be the number of ways to tile the grid so that:

  • All cells from cell (0,0)(0, 0) to cell (i,j1)(i, j - 1) are covered.
  • All cells from cell (i+1,j)(i + 1, j) to cell (N1,M1)(N - 1, M - 1) are empty.
  • maskmask represents whether each of the remaining NN cells are empty, with the kk-th bit corresponding to the cell in column kk.

For example, the following state would contribute toward dp[1][3][101002]\texttt{dp}[1][3][10100_2]: broken profile state example

The answer will be dp[N1][M1][0]\texttt{dp}[N - 1][M - 1][0].

We now have three cases when calculating dp[i][j][mask]\texttt{dp}[i][j][mask]:

  • The ii-th bit of the mask is 0, meaning that cell (i,j)(i, j) is covered.
    • Case 1: we used a vertical tile to cover it.
      • Cell (i,j1)(i, j - 1) must have been empty, so there are dp[i1][j][mask2i]\texttt{dp}[i - 1][j][mask \oplus 2^i] ways to do this.
    • Case 2: we used a horizontal tile to cover it.
      • This is only possible if i>0i > 0.
      • Cells (i1,j)(i - 1, j), (i1,j1)(i - 1, j - 1), and (i,j1)(i, j - 1) must have been covered, so there are dp[i1][j][mask2i1]\texttt{dp}[i - 1][j][mask \oplus 2^{i - 1}] ways to do this.
      • This corresponds to if (i && !(mask & (1 << i)) && !(mask & (1 << i - 1))) in the code below.
  • The ii-th bit of the mask is 1, meaning that cell (i,j)(i, j) is empty.
    • Cell (i,j1)(i, j - 1) must have been covered, so there are dp[i1][j][mask2i]\texttt{dp}[i - 1][j][mask \oplus 2^i] ways to do this.
      • This is the same as case 1 of when the ii-th bit of the mask is 0, so we handle them simultaneously in the code below.

Note that the indices we need to use may become negative and will thus require wrapping. To simplify calculations and bypass this, simply drop the first two dimensions of the DP array, as dp[i][j]\texttt{dp}[i][j] depends only on dp[i1][j]\texttt{dp}[i - 1][j].



#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
int dp[1 << 10][2];
int main() {
int n, m;

Broken profile DP Problems

StatusSourceProblem NameDifficultyTags
Show TagsBroken Profile
Show TagsBroken Profile
PlatVery Hard
Show TagsBroken Profile

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