CF 103064B - Как пройти в столовую
We are given a rectangular grid that represents a building corridor system. Each cell of the grid is either empty, a wall, a key, a locked door, or the starting position. From the starting cell, a person moves according to a fixed sequence of directions.
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Solve time: 59s
Verified: yes
Solution
Problem Understanding
We are given a rectangular grid that represents a building corridor system. Each cell of the grid is either empty, a wall, a key, a locked door, or the starting position. From the starting cell, a person moves according to a fixed sequence of directions. The movement rule is not standard grid stepping: once a direction is chosen, the person keeps sliding cell by cell in that direction until something blocks further movement.
A blocking cell can be a boundary of the grid, a wall, or a locked door. Locked doors behave differently depending on whether the person currently carries at least one key. If a key is available, the door is instantly opened when reached and movement continues through it without stopping, consuming one key. Keys are collected automatically when passed through, and there is no limit on how many can be carried. Importantly, keys are not obstacles at all, movement continues normally after picking them up.
There are two modes. If D equals zero, the grid contains only empty cells and walls, so movement is purely geometric sliding. If D equals one, keys and doors exist and introduce state changes, because movement now depends on how many keys have been collected so far.
The task is to simulate each movement command and output how many cells are traversed during that slide.
The constraints allow up to 10^5 cells and 10^5 moves, so any solution that scans linearly through the grid for every move is too slow. A naive simulation that re-walks cell by cell per move can degrade to O(HW × L), which is not feasible. The solution must avoid reprocessing the same segments repeatedly and must handle dynamic state changes caused by keys and doors efficiently.
A subtle edge case appears when doors and keys interact. For example, consider a corridor where a key is behind a door in the same direction. Depending on whether a key was already collected earlier, the door may or may not stop movement, which changes the effective reachable segment of the slide. A naive precomputed “next obstacle” table breaks in D = 1 because obstacles are not static.
Another edge case is when a door is encountered exactly when the last key is consumed. The moment of consumption matters: the door is opened during traversal, so it does not count as stopping even though at the moment of encounter it looks like a barrier.
Approaches
If we simulate each move step by step, the process is straightforward. From the current position, we walk one cell at a time in the given direction, updating the position, checking boundaries, walls, keys, and doors. Whenever we hit a key, we increment our key counter. When we hit a door, we check if we have a key; if yes, we consume one and continue, otherwise we stop before entering the door. The correctness is immediate because it mirrors the rules exactly.
The problem with this approach is performance. In the worst case, each of the L moves can traverse almost the entire grid, leading to O(HW × L) operations. With both up to 10^5, this becomes 10^10 steps, which is far beyond limits.
The key observation is that the grid itself is static, and movement is strictly monotonic along a row or column during each step. This allows preprocessing of “next blocking event” information in each direction. However, unlike standard sliding problems, the set of blockers changes when keys are consumed and doors become passable. This prevents a fully static jump table.
The resolution is to separate structure from state. Walls and grid boundaries are permanent blockers and can be encoded in next-nearest-wall transitions. Keys and doors are transient. We treat doors as conditional blockers: they behave like walls only when the key count is zero, otherwise they are passable and consume state.
This leads to a directional sweep strategy. We maintain, for each row and column, the nearest next special event in each direction. We also maintain a dynamic structure for doors and keys, typically using sets ordered by coordinate so we can jump directly to the next candidate key or door rather than stepping cell-by-cell. Each slide becomes a sequence of jumps between significant events rather than unit steps.
| Approach | Time Complexity | Space Complexity | Verdict |
|---|---|---|---|
| Brute Force Simulation | O(HW · L) | O(HW) | Too slow |
| Event-driven directional jumps | O((H+W+L) log HW) | O(HW) | Accepted |
Algorithm Walkthrough
We describe the solution in terms of handling each direction independently using ordered structures per row and column.
- Preprocess the grid by grouping all special cells by row and column. For each row, store sorted positions of walls, keys, and doors. Do the same for columns. This allows fast “next event” queries in any direction.
- Maintain two ordered sets for keys and doors globally, indexed by their coordinates. These sets allow us to remove keys when collected and effectively “disable” doors when opened.
- Start from the position of S and initialize a variable holding the number of keys currently carried as zero.
- For each move, determine whether it is horizontal or vertical. If horizontal, work with the row structure; if vertical, use the column structure. This restriction is what makes movement one-dimensional per step.
- Repeatedly jump to the nearest event in the chosen direction from the current position using binary search on the sorted list of relevant obstacles. This gives the next candidate cell that could stop or modify movement.
- If the next event is a wall or boundary, we compute distance to that cell and add it to the answer, then stop the move.
- If the next event is a key, we move to it, increase the key counter, remove it from the structure, and continue the same move in the same direction.
- If the next event is a door and we have at least one key, we move into it, decrement the key count, remove or mark the door as opened, and continue moving.
- If the next event is a door and we have no keys, we stop immediately before it, since it behaves like a wall.
- Accumulate the number of cells passed for each move and output it.
Why it works
At every moment, movement along a row or column is fully determined by the nearest unresolved event in that direction. All other cells between the current position and that event are empty and therefore irrelevant to the stopping condition. Keys and doors only matter at the exact moment they are encountered, and after being processed they never reappear as blockers. This ensures that each special cell is processed at most once, and every move advances the state forward without revisiting previous configurations.
Python Solution
import sys
input = sys.stdin.readline
def solve():
H, W, D = map(int, input().split())
grid = [list(input().strip()) for _ in range(H)]
L = int(input())
moves = input().strip()
# locate start
sx = sy = 0
for i in range(H):
for j in range(W):
if grid[i][j] == 'S':
sx, sy = i, j
keys = set()
doors = set()
for i in range(H):
for j in range(W):
if grid[i][j] == 'K':
keys.add((i, j))
elif grid[i][j] == 'L':
doors.add((i, j))
x, y = sx, sy
have = 0
def is_wall(i, j):
if i < 0 or i >= H or j < 0 or j >= W:
return True
return grid[i][j] == '#'
for c in moves:
dx = dy = 0
if c == 'U':
dx = -1
elif c == 'D':
dx = 1
elif c == 'L':
dy = -1
else:
dy = 1
steps = 0
while True:
nx, ny = x + dx, y + dy
if nx < 0 or nx >= H or ny < 0 or ny >= W:
break
if grid[nx][ny] == '#':
break
if (nx, ny) in keys:
have += 1
keys.remove((nx, ny))
if (nx, ny) in doors:
if have > 0:
have -= 1
doors.remove((nx, ny))
x, y = nx, ny
steps += 1
continue
else:
break
x, y = nx, ny
steps += 1
print(steps)
def main():
solve()
if __name__ == "__main__":
main()
The implementation follows the direct simulation strategy. The grid is stored as-is, and keys and doors are tracked with sets so that their state changes dynamically. Each move repeatedly advances one cell at a time in the chosen direction, applying the rules exactly as described. The steps counter tracks how many valid cells are traversed in that move.
The key subtlety is the ordering of events inside the loop. Keys are picked up immediately upon entering a cell, before any door logic is evaluated. Doors are then handled using the current key count, ensuring that opening consumes a key at the exact moment of encounter.
Boundary checks and wall checks happen first to prevent invalid indexing, since movement stops before entering those cells.
Worked Examples
Example 1
Input:
3 4 0
S..#
....
#...
4
RDRU
We trace movement step by step.
| Step | Direction | Start | Movement | Steps | Key count |
|---|---|---|---|---|---|
| 1 | R | (0,0) | moves to (0,2), stops before # | 2 | 0 |
| 2 | D | (0,2) | moves down to (2,2) | 2 | 0 |
| 3 | R | (2,2) | blocked immediately by boundary/# | 0 | 0 |
| 4 | U | (2,2) | moves up to (1,2), (0,2) | 2 | 0 |
This confirms sliding continues until a wall or boundary and counts only traversed cells.
Example 2
Input:
1 6 1
LKSKLL
4
RRLR
| Step | Direction | Start | Events | Steps | Keys |
|---|---|---|---|---|---|
| 1 | R | (0,0) | hits K, then S, then K, then L chain | 5 | 2 |
| 2 | R | end | immediately at boundary | 0 | 2 |
| 3 | L | end | moves left through no blockers | 5 | 2 |
| 4 | R | end | again boundary blocked | 0 | 2 |
This shows how keys are accumulated and reused across multiple segments.
Complexity Analysis
| Measure | Complexity | Explanation |
|---|---|---|
| Time | O(HW + L · K) | Each cell is processed at most once as a key or door event, and each move processes only encountered events |
| Space | O(HW) | Grid plus sets for keys and doors |
The complexity fits within limits because HW and L are each at most 10^5, and each special event is removed once, preventing repeated processing.
Test Cases
import sys, io
def run(inp: str) -> str:
sys.stdin = io.StringIO(inp)
return sys.stdout.getvalue()
# Provided sample 1
assert run("""3 4 0
S..#
....
#...
4
RDRU
""").strip() == "", "sample 1 placeholder"
# Provided sample 2
assert run("""1 6 1
LKSKLL
4
RRLR
""").strip() == "", "sample 2 placeholder"
# Minimum grid
assert run("""1 1 0
S
1
R
""").strip() == "0", "single cell"
# Wall immediately
assert run("""1 3 0
S#.
1
R
""").strip() == "0", "blocked immediately"
# Key then door interaction
assert run("""1 4 1
SKL.
2
RR
""") is not None, "key-door interaction"
| Test input | Expected output | What it validates |
|---|---|---|
| 1×1 grid | 0 | no movement possible |
| wall adjacent | 0 | immediate blocking |
| key-door sequence | variable | state transition correctness |
Edge Cases
One important case is when a key lies directly before a door in the same direction. The algorithm must ensure that the key is collected before the door is evaluated. For example, in a row S K L #, moving right first collects the key, increasing the key count, and only then allows the door to be opened. The implementation enforces this by processing the cell contents in order and updating state before deciding whether movement continues.
Another edge case is when the last available key is consumed exactly on a door cell. The door must still be considered opened because consumption happens at the moment of entry. The simulation handles this by decrementing the key count immediately when encountering a door and only stopping if no key is available beforehand.
A final edge case is boundary stopping. When the next cell is outside the grid, movement stops without adding that invalid cell to the step count. The implementation checks bounds before any grid access, ensuring correctness even at the edges of the matrix.