**Subtask 1** ($K \leq 5$)

We can do this via bitmask DP. Our states will be pairs $(\mu, r)$ where $\mu$ is a bitmask of length $K$ and $r$ is a room. The bitmask $\mu$ represents all sequences of button presses such that iff the $j$th bit in $\mu$ is on, Bessie has pressed button $j$ at least once and has not pressed any button with a higher number than $j$ since the last time she pressed button $j$. The state $(\mu, r)$ then represents all sequences of button presses and rooms beginning in room $s$ by pressing button $b_s$, ending in room $r$, and satisfying the conditions to correspond to mask $\mu$.

A transition in our DP is then moving to another (possibly the same) room as well as pressing another button. Bessie can move to room $r’$ iff there is an edge from $r$ to $r’$. Additionally, Bessie can press button $j$ iff the $j$th bit in the current mask $\mu$ is off — otherwise, she would press button $j$ twice without pressing a button with a higher number in between.

We then must also consider how pressing a button changes $\mu$. It clearly sets the $j$th bit to be on. Additionally, all lower bits are set off now that a higher button has been pressed. All higher bits are unaffected. It is important to note that the way this modification works implies that we always increase the numerical value of our bitmask when transitioning, meaning that there are no circular relations in our DP and that we should consider bitmasks in order of increasing numerical value.

Our DP then has $O(2^KN)$ states, from each of which we perform $O(KN)$ transitions, meaning that our DP runs in $O(K2^KN^2)$.

To compute the answer for each query, we simply take the sum of the DP values of $(t, \mu)$ over all bitmasks $\mu$ such that the $b_t$th bit is on and all lower bits are off, as these correspond precisely to the sequences where we end by pressing button $b_t$. This takes $O(2^K)$ time, so that our final computations take $O(2^KQ)$ time overall and thus our overall solution takes $O(2^K(KN^2 + Q))$.

**Subtask 2** ($b_s = K - 1$ and $b_t = K$ for each query)

First note that as Bessie always presses button $K$ at the end, she cannot press button $K$ before then, as there would be no higher button that she could press in between (as button $K$ is the highest button). This also means that she cannot press button $K - 1$ after the beginning as she could never press button $K$, the only higher button. Thus, any sequence of button presses that Bessie takes, excluding the first and last button, uses buttons with numbers at most $K - 2$.

We can then solve this using DP. $dp_{a, b, x}$ will be the number of sequences of rooms and button presses which start at room $a$, end at room $b$, and only involve pressing buttons with numbers at most $x$. We can compute transitions as follows. First, we add all of $dp_{a, b, x - 1}$ to $dp_{a, b, x}$, as a sequence only using buttons with numbers at most $x - 1$ will clearly also only use buttons with numbers at most $x$.

We then consider the case where we do press button $x$. Note that as we never press any higher buttons, we must only press button $x$ once. Let’s then fix the room $r$ that Bessie is in when she presses button $x$, and consider in isolation the part of the sequence before pressing button $x$ (the left side) and after pressing button $x$ (the right side). The left side, if it is not empty, starts at some room $a$, ends at some room $b$ such that there is an edge from $b$ to $r$, and only presses buttons with number at most $x - 1$. Similarly, the right side, if it is not empty, starts at some room $c$ such that there is an edge from $r$ to $c$, ends at some room $d$, and only presses buttons with number at most $x - 1$.

The number of left sides is then $dp_{a, b, x - 1}$ for each $a, b$ and the number of right sides is $dp_{c, d, x - 1}$ for each $c, d$. We only care about the endpoints $a$ and $d$, so for each $a$, we compute the sum over all $dp_{a, b, x - 1}$ such that there is an edge from $b$ to $r$, and similarly for each $d$. Note that we should account for the left side possibly being empty by adding $1$ to our calculation for $a = r$, as in that case our overall left endpoint will just be $r$ (and similarly for the right side). Finally, for each $a, d$, we can simply multiply these quantities together and add the result to $dp_{a, d, x}$.

Each of these three computations is $O(N^2)$, so as we do them for each maximum number $x$ and “middle” room $r$, our DP computation is $O(N^3K)$ overall.

Given this DP, for each query to compute the answer, given that we want to start at room $s$ and end at room $t$, we loop through all second rooms $a$ (such that there is an edge from $s$ to $a$) and second-to-last rooms $b$ (such that there is an edge from $b$ to $t$) and add $dp_{a, b, K - 2}$ to our answer. This is $O(N^2)$ for each query, making our overall algorithm $O(N^3K + QN^2)$.

**Subtask 3** ($N, K, Q \leq 20$)

We can solve this subtask by modifying our solution for the previous subtask. First note that our constraints are now low enough to allow computing our $O(N^3K)$ DP for each query. Given this, we can simply modify our DP as follows. $dp_{a, b, k}$ will be defined as usual, except that $a$ can also take a special value $\alpha$ which will indicate that we are counting sequences which start at room $s$ but must also start by pressing button $b_s$. Similarly, $b$ can take a special value $\beta$ which will indicate that we are counting sequences which end at room $t$ but must also end by pressing button $b_t$.

These special values $\alpha$ and $\beta$ then essentially function as normal rooms during our DP. There are only two differences: ones is that in our above loops through all $a, b$ and $c, d$, the rooms $b$ and $c$ are not allowed to be $\alpha$ or $\beta$ as that would lead to overcounting; the other is that we must properly account for the left side being empty by also adding $1$ to our calculation for $a = \alpha$ if we are currently at $r = s$ and $x = b_s$, and similarly for the right side.

Our answer is then simply $dp_{\alpha, \beta, K}$. These modifications do not affect the asymptotic runtime of our DP, but we do now do it for each query, giving the overall algorithm a runtime of $O(QN^3K)$.

**Subtask 4** (original constraints)

Our solution for this is essentially the same as for the previous subtask, noting that we don’t actually really need to compute this DP for each query. We can compute this DP just once by, instead of having the two special values $\alpha$ and $\beta$, having values $\alpha_j$ and $\beta_j$ for each $1 \leq j \leq Q$ such that $\alpha_j$ corresponds to $(s, b_s)$ for the $j$th query and $\beta_j$ corresponds to $(t, b_t)$ for the $j$th query. The answer to the $j$th query is then $dp_{\alpha_j, \beta_j, K}$.

Considering the additional states, the runtime of our algorithm is now $O(N(N + Q)^2K)$, which is still fast enough.

Danny's code:

import java.util.Scanner; public class LevelGraph { public static final long MOD = 1000000007; public static void main(String[] args) { Scanner in = new Scanner(System.in); int n = in.nextInt(); int k = in.nextInt(); int q = in.nextInt(); boolean[][] adj = new boolean[n][n]; for (int a = 0; a < n; a++) { String line = in.next(); for (int b = 0; b < n; b++) { adj[a][b] = line.charAt(b) == '1'; } } int[] xs = new int[q]; int[] froms = new int[q]; int[] ys = new int[q]; int[] tos = new int[q]; for (int j = 0; j < q; j++) { xs[j] = in.nextInt() - 1; froms[j] = in.nextInt() - 1; ys[j] = in.nextInt() - 1; tos[j] = in.nextInt() - 1; } long[][][] dp = new long[k][n + q][n + q]; for (int h = 0; h < k; h++) { if (h > 0) { for (int a = 0; a < n + q; a++) { for (int b = 0; b < n + q; b++) { dp[h][a][b] = dp[h - 1][a][b]; } } } for (int c = 0; c < n; c++) { long[] left = new long[n + q]; left[c] = 1; for (int j = 0; j < q; j++) { if (h == xs[j] && froms[j] == c) { left[n + j] = 1; } } if (h > 0) { for (int a = 0; a < n + q; a++) { for (int b = 0; b < n; b++) { if (adj[b][c]) { left[a] += dp[h - 1][a][b]; left[a] %= MOD; } } } } long[] right = new long[n + q]; right[c] = 1; for (int j = 0; j < q; j++) { if (h == ys[j] & tos[j] == c) { right[n + j] = 1; } } if (h > 0) { for (int a = 0; a < n; a++) { for (int b = 0; b < n + q; b++) { if (adj[c][a]) { right[b] += dp[h - 1][a][b]; right[b] %= MOD; } } } } for (int a = 0; a < n + q; a++) { for (int b = 0; b < n + q; b++) { dp[h][a][b] += left[a] * right[b]; dp[h][a][b] %= MOD; } } } } for (int j = 0; j < q; j++) { System.out.println(dp[k - 1][n + j][n + j]); } } }