Editorial: Seating Arrangement (Easy Version)
Every table starts empty. The first person to sit at a table is the opener; all subsequent people at that table are fillers. The personality rules translate directly:
| Role | Allowed types |
|---|---|
| Opener (first at table) | I or A |
| Filler (subsequent at table) | E or A |
A table with $s$ seats provides 1 opener slot and $s - 1$ filler slots. The ordering constraint says: an opener’s position in the queue must come before all fillers at the same table.
Since I’s can only open and E’s can only fill, the interesting decision is: when should an A open a new table vs. join an occupied one?
A natural greedy idea: “if a future I needs an empty table, don’t waste one on A.” Formally, let $f_I$ = number of future I’s and $e$ = empty table count; when $f_I \ge e$, have A fill instead. This rule is wrong.
Counterexample: $x = 2$, $s = 3$, queue I A E E E I E.
- A fills (greedy rule since $f_I = 1 = e$): I seats A in T1 (now full). E(2) sits, E(3) and E(4) are kicked. I opens T2. E(6) sits. Total: 5.
- A opens T2: four filler slots available. E(2), E(3), E(4) all sit. I(5) is kicked. E(6) sits. Total: 6.
Opening creates $s - 1 = 2$ extra filler slots, letting the three E’s between positions 1 and 5 sit. The greedy argument “displacing one I is net 0” misses these E’s. The correct approach is DP.
$$dp[e] = \text{maximum people seated so far with exactly } e \text{ empty tables remaining.}$$
Use a rolling 1-D array of size $x + 1$, processing people left to right.
Given $e$ and $ans = dp[e]$, the remaining filler capacity across all opened tables is:
$$\text{filler\_slots} = (x - e) \cdot s - ans.$$Derivation:
- Tables opened $= x - e$.
- Openers seated $= x - e$.
- Fillers seated $= ans - (x - e)$.
- Filler capacity $= (x - e)(s - 1)$.
- Remaining $= (x-e)(s-1) - (ans - (x-e)) = (x - e) \cdot s - ans$.
This formula is the heart of the solution: it eliminates any need to track individual table occupancies.
For each person at position $i$ (let $fs = (x-e) \cdot s - ans$):
| Person | Action | New state |
|---|---|---|
| I | kick | $(e,\ ans)$ |
| I | open (if $e > 0$) | $(e - 1,\ ans + 1)$ |
| E | fill (if $fs > 0$) | $(e,\ ans + 1)$ |
| E | kick (if $fs = 0$) | $(e,\ ans)$ |
| A | fill (if $fs > 0$) | $(e,\ ans + 1)$ |
| A | open (if $e > 0$) | $(e - 1,\ ans + 1)$ |
| A | kick (if $e = 0$, $fs = 0$) | $(e,\ ans)$ |
For each reachable $(e, ans)$, all applicable transitions are applied and the new state takes the maximum.
Note: filling E is always at least as good as kicking E (when $fs > 0$), so filling is always taken. For A, both fill and open are explored; the DP chooses whichever leads to the better future.
Answer: $\displaystyle\max_{0 \le e \le x} dp[n][e]$.
$O(n \cdot x)$ per test case. With $n, x \le 3000$ and $\sum n \le 3000$, the total work is at most $3000 \times 3000 = 9 \times 10^6$.
$x = 2$, $s = 3$, queue I A E E E I E ($n = 7$). Expected: 6.
Trace the key states (omitting unreachable cells):
| $i$ | char | $e = 2$ | $e = 1$ | $e = 0$ |
|---|---|---|---|---|
| init | — | 0 | — | — |
| 0 I | I opens | 0 | 1 | — |
| 1 A | fills T1 OR opens T2 | 0 | 2 | 2 |
| 2 E | fills | 0 | 3 | 3 |
| 3 E | fills | 0 | 3 | 4 |
| 4 E | fills | 0 | 3 | 5 |
| 5 I | opens | 0 | 3 | 5 |
| 6 E | fills | 0 | 3 | 6 |
Answer: $\max(0, 3, 6) = \mathbf{6}$.
The state $e = 0, ans = 6$ comes from the path: I opens T1, A opens T2 (at step 1 the $e=0$ branch), E fills three times, I is kicked (e=0 at step 5), E fills. Exactly 6 people seated.
$x = 2$, $s = 2$, queue EIAIE. Expected: 4.
| $i$ | char | $e=2$ | $e=1$ | $e=0$ |
|---|---|---|---|---|
| init | — | 0 | — | — |
| 0 E | kick ($fs=0$) | 0 | — | — |
| 1 I | opens | 0 | 1 | — |
| 2 A | fills T1 or opens T2 | 0 | 2 | 2 |
| 3 I | opens | 0 | 2 | 3 |
| 4 E | fills | 0 | 2 | 4 |
Answer: $\max(0, 2, 4) = \mathbf{4}$. ✓