## G - Digits on Grid Editorial by en_translator

Consider the following undirected graph \(G\):

- It has \(H\) vertices \(a_1, a_2, \ldots, a_H\) corresponding to each row and \(W\) vertices \(b_1, b_2, \ldots, b_W\) corresponding to each column, for a total of \((H+W)\) vertices.
- For each pair of integers \((i, j)\) such that \(1 \leq i \leq H\) and \(1 \leq j \leq W\), there is an undirected edge labeled \(c_{i, j}\) between Vertex \(a_i\) and Vertex \(b_j\).
- There are no other vertices or edges.

Let \(\mathcal{R}\) be the set of vertices corresponding to the rows, and \(\mathcal{C}\) be the set of vertices corresponding to the columns. That is,

\(\mathcal{R} := \lbrace a_1, a_2, \ldots, a_H \rbrace\)

\(\mathcal{C} := \lbrace b_1, b_2, \ldots, b_W \rbrace\)

Note that \(G\) is a bipartite graph whose vertices are divided into \(\mathcal{R}\) and \(\mathcal{C}\).

This problem can be rephrased as follows.

Starting from a vertex belonging to \(\mathcal{R}\), we repeat “traveling from the current vertex to one of the adjacent vertices along the edge between them” \(2N\) times. Find the number of different sequences of labels \(d_1d_2\ldots d_{2N}\) that can be obtained by concatenating the labels on the traversed edges in order.

We will solve this problem with Dynamic Programming (DP). If we define DP table as

\(\mathrm{dp}[k][v] := \)(the number of different sequences of labels \(d_1d_2\ldots d_k\) corresponding to the first \(k\) moves ending with vertex \(v\)),

the same sequences of label can be counted multiple times, so we cannot obtain the correct answer.
This is because, for a sequence of label \(d_1d_2\ldots d_k\), there may be in general multiple possible “current vertex \(v\)” right after traveling through the label sequence.

In order to deal with this issue, instead of single “terminal vertex \(v\)”, we use “the set of all possible terminal vertices as the results of the label sequence” for each label sequence \(d_1d_2\ldots d_k\), so that the object and the sequence of labels correspond one-by-one, and thus counted without any duplicates.

In other words, we fill the DP table defined as:

\(\mathrm{dp}[k][S] := \)（The number of different sequences of labels of length \(k\) such that the set of possible terminal vertices after traversing those labels in the sequences is \(S\)).

Since graph \(G\) is a bipartite graph,

- after even number of moves, the current vertex is limited to those in \(\mathcal{R}\);
- after odd number of moves, the current vertex is limited to those in \(\mathcal{C}\).

Therefore,

when considering \(\mathrm{dp}[k][S]\), it is sufficient to consider

- only \(S\) such that \(S \subseteq \mathcal{R}\) when \(k\) is event;
- only \(S\) such that \(S \subseteq \mathcal{C}\) when \(k\) is odd.

Therefore, the number of states of DP to consider is \(\mathrm{O}(N(2^H + 2^W))\).

We will now describe the specific steps of DP.

First, as the initial state of DP, we consider the case of \(k = 0\), that is, \(\mathrm{dp}[0][\ast]\). Since the initial position before the first move can be chosen from arbitrary vertex belonging to \(\mathcal{R}\), so

\(\mathrm{dp}[0][\mathcal{R}] = 1\);

\(\mathrm{dp}[0][S] = 0, \,\,\text{for}\,\,S \neq \mathcal{R}\).

Next, we consider the transitions of the DP.

Consider a sequence of labels \(d_1 d_2 \ldots d_k\) of length \(k\). Also, let \(S\) be the set of possible vertices right after traversing through these labels.

Then, for a label sequence \(d_1 d_2 \ldots d_kc\) where digit \(c\) is appended to the original one, let \(S'_c\) be the set of all possible vertices right after traversing through them; then

- If \(k\) is even, then \(S'_c = \lbrace b_j :\) there exists \(i \in S\) such that \(c_{i, j} = c\);
- If \(k\) is odd, then \(S'_c = \lbrace a_i :\) there exists \(j \in S\) such that \(c_{i, j} = c\).

Based on these fact, it is sufficient to perform the following nine transitions:

- \(\mathrm{dp}[k+1][S'_1] \leftarrow \mathrm{dp}[k+1][S'_1] + \mathrm{dp}[k][S]\)
- \(\mathrm{dp}[k+1][S'_2] \leftarrow \mathrm{dp}[k+1][S'_2] + \mathrm{dp}[k][S]\)
- \(\mathrm{dp}[k+1][S'_3] \leftarrow \mathrm{dp}[k+1][S'_3] + \mathrm{dp}[k][S]\)
- \(\cdots\)
- \(\mathrm{dp}[k+1][S'_9] \leftarrow \mathrm{dp}[k+1][S'_9] + \mathrm{dp}[k][S]\)

each of which corresponds to the nine cases where the final digit \(c\) are \(1\) to \(9\), respectively.

By filling the DP table with the initializations and transitions above, the answer for this problem is obtained as

\(\displaystyle\sum_{S \in \mathcal{R}} \mathrm{dp}[2N][S] - \mathrm{dp}[2N][\emptyset]\).

Therefore, the problem has been solved in a total of \(O(N(2^H+2^W))\) time.

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