CF 106270C - Gas Reservoir
We are given a 3D grid representing underground space. Each cell is either rock or gas, and every gas cell contains exactly one unit of gas.
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Solve time: 59s
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Solution
Problem Understanding
We are given a 3D grid representing underground space. Each cell is either rock or gas, and every gas cell contains exactly one unit of gas. Gas cells connect in 3D using face adjacency, so any connected component of gas forms a “reservoir” that is indivisible: if you touch one cell of it, the whole component is considered collected.
We are allowed to pick a single vertical drilling column at some surface position $(x, y)$. The drill goes straight down through all depths, and whenever it passes through any gas cell, it triggers collection of the entire reservoir that contains that cell. If the same reservoir is encountered multiple times along the same column, it is only counted once. The goal is to choose one column so that the total size of all distinct reservoirs touched by that column is maximized.
So the problem is not about counting gas cells locally in a column. Instead, a column acts as a “trigger set” for whole connected components in 3D.
The grid dimensions are up to $50 \times 50 \times 50$, so the total number of cells is at most 125,000 per test case. This size allows linear or near-linear traversals over the grid, but rules out anything quadratic over all cells or cubic per column.
A naive mistake would be to treat each depth slice independently and sum gas cells per column, or to count gas per layer. That fails because reservoirs span multiple columns and depths.
For example, consider a single large connected component shaped like a hollow 3D blob touching multiple $(x,y)$ columns. If you pick one column that intersects it, you get the entire component, not just the cells in that column. A local counting strategy would severely underestimate this.
Another subtle issue is duplicate counting across depth. If a reservoir appears in multiple z-levels under the same $(x,y)$, it must still only be counted once. This makes it necessary to work with connected components explicitly.
Approaches
A direct brute-force approach is easy to describe. For every surface coordinate $(x,y)$, we simulate drilling down that column and collect all gas cells encountered. Every time we hit a gas cell that belongs to a reservoir not yet counted for that column, we would flood-fill the entire component and add its size.
This is correct, but inefficient if implemented naively. The expensive part is repeatedly exploring connected components. In the worst case, every gas cell could be its own component or belong to a large component that is rediscovered many times across columns. If we run a flood fill for every column independently, we can end up exploring up to $O(XYZ)$ per column, giving $O((XY)(XYZ))$, which is far too large.
The key observation is that reservoirs are global objects. Once we identify each connected component of gas and compute its size, every cell simply becomes a pointer to that component. Then the problem reduces to selecting a set of components that intersect a given column. Instead of recomputing connectivity per column, we precompute all components once, assign each cell a component id, and store each component’s size.
After that, for each column $(x,y)$, we only need to check which distinct component ids appear along depth $z$, and sum their sizes once. This transforms the problem into a two-phase process: global preprocessing of connected components, then local aggregation per column.
Because each cell belongs to exactly one component, and we only scan each column once, the total work becomes linear in the grid size.
| Approach | Time Complexity | Space Complexity | Verdict |
|---|---|---|---|
| Brute Force (DFS per column) | $O((XY)(XYZ))$ | $O(XYZ)$ | Too slow |
| Component labeling + column aggregation | $O(XYZ)$ | $O(XYZ)$ | Accepted |
Algorithm Walkthrough
We treat the grid as a 3D graph where each gas cell is a node, and edges exist between face-adjacent gas cells.
- Traverse every cell in the grid. Whenever we find a gas cell that has not been assigned to any component, we start a flood fill (BFS or DFS) from it. This flood fill marks all reachable gas cells as part of the same component and assigns them a component id. During this traversal, we also count how many cells belong to this component. This count is stored as the component’s size.
- After this preprocessing step, every gas cell knows which component it belongs to, and every component knows its total gas volume. This step ensures we have fully compressed the 3D structure into disjoint sets.
- For each surface coordinate $(x, y)$, we scan vertically from $z = 0$ to $z = Z-1$. We maintain a boolean set or timestamp array to ensure that we do not count the same component twice for a single column.
- Whenever we encounter a gas cell at $(x,y,z)$, we look up its component id. If this component has not yet been counted for this column, we add its precomputed size to the column sum and mark it as visited for this column.
- After processing all depths for a column, we compare its total with the global maximum and keep the best result.
- Finally, output the maximum over all columns.
The crucial design choice is that the “visited” tracking is reset per column, not globally, because each column independently selects components.
Why it works
The correctness rests on two structural properties. First, every gas cell belongs to exactly one connected component, so grouping is well-defined and disjoint. Second, a column collects a component if and only if it contains at least one cell from that component. Therefore, summing component sizes exactly matches the total gas collected, provided each component is counted once per column. Since the algorithm enforces that constraint using a per-column visited set, no component is double counted and no reachable component is missed.
Python Solution
import sys
input = sys.stdin.readline
from collections import deque
def solve():
T = int(input())
for _ in range(T):
line = input().strip()
while line == "":
line = input().strip()
X, Y, Z = map(int, line.split())
grid = [[[None]*Y for _ in range(X)] for _ in range(Z)]
for z in range(Z):
blank = input().strip()
while blank == "":
blank = input().strip()
for x in range(X):
row = input().strip()
grid[z][x] = list(row)
comp = [[[-1]*Y for _ in range(X)] for _ in range(Z)]
comp_size = []
dirs = [(1,0,0), (-1,0,0), (0,1,0), (0,-1,0), (0,0,1), (0,0,-1)]
cid = 0
for z in range(Z):
for x in range(X):
for y in range(Y):
if grid[z][x][y] == '.' and comp[z][x][y] == -1:
q = deque()
q.append((z,x,y))
comp[z][x][y] = cid
cnt = 0
while q:
cz, cx, cy = q.popleft()
cnt += 1
for dz, dx, dy in dirs:
nz, nx, ny = cz+dz, cx+dx, cy+dy
if 0 <= nz < Z and 0 <= nx < X and 0 <= ny < Y:
if grid[nz][nx][ny] == '.' and comp[nz][nx][ny] == -1:
comp[nz][nx][ny] = cid
q.append((nz,nx,ny))
comp_size.append(cnt)
cid += 1
best = 0
seen = [0]*len(comp_size)
stamp = 1
for x in range(X):
for y in range(Y):
total = 0
stamp += 1
local_seen = seen
for z in range(Z):
if grid[z][x][y] == '.':
c = comp[z][x][y]
if local_seen[c] != stamp:
local_seen[c] = stamp
total += comp_size[c]
best = max(best, total)
print(best)
if __name__ == "__main__":
solve()
The first phase builds connected components using BFS in the full 3D grid. Each gas cell is assigned a component id exactly once, and the BFS ensures all face-connected gas is grouped together. The size array stores how many cells each reservoir contains.
The second phase iterates over all $(x,y)$ surface positions. For each column, we scan downward and check which component ids appear. The timestamp trick avoids allocating a fresh set per column, which keeps memory and time efficient. Each component is added at most once per column.
A subtle point is that we only trigger component collection when we see a gas cell, but since every gas cell has a valid component id, this fully captures all reservoirs intersecting the column.
Worked Examples
Example 1
Consider a tiny grid where two separate gas components exist and only one column intersects both.
| Step | (x,y) column | z scan | encountered components | total |
|---|---|---|---|---|
| 1 | (0,0) | z=0..Z | {A} | 3 |
| 2 | (1,0) | z=0..Z | {B} | 5 |
| 3 | (1,1) | z=0..Z | {A,B} | 8 |
This demonstrates that the algorithm correctly merges contributions only when a column intersects multiple reservoirs.
Example 2
A single large connected reservoir spanning the entire grid.
| Step | (x,y) column | z scan | encountered components | total |
|---|---|---|---|---|
| 1 | any column | z=0..Z | {A} | 125 |
Every column touching the reservoir gets the full size because BFS already merged all cells into one component. The column selection no longer matters structurally.
This confirms that connectivity is fully captured in preprocessing and column logic only aggregates.
Complexity Analysis
| Measure | Complexity | Explanation |
|---|---|---|
| Time | $O(XYZ)$ | Each cell is visited once in BFS and once in column scanning |
| Space | $O(XYZ)$ | Storage for grid, component ids, and sizes |
The constraints allow up to 125,000 cells per test case, and at most 10 test cases, so the solution comfortably fits within limits with linear traversal and constant-time checks per cell.
Test Cases
import sys, io
def run(inp: str) -> str:
sys.stdin = io.StringIO(inp)
from collections import deque
def solve():
T = int(input())
for _ in range(T):
line = input().strip()
while line == "":
line = input().strip()
X, Y, Z = map(int, line.split())
grid = [[[None]*Y for _ in range(X)] for _ in range(Z)]
for z in range(Z):
blank = input().strip()
while blank == "":
blank = input().strip()
for x in range(X):
row = input().strip()
grid[z][x] = list(row)
comp = [[[-1]*Y for _ in range(X)] for _ in range(Z)]
comp_size = []
dirs = [(1,0,0), (-1,0,0), (0,1,0), (0,-1,0), (0,0,1), (0,0,-1)]
cid = 0
for z in range(Z):
for x in range(X):
for y in range(Y):
if grid[z][x][y] == '.' and comp[z][x][y] == -1:
q = deque()
q.append((z,x,y))
comp[z][x][y] = cid
cnt = 0
while q:
cz, cx, cy = q.popleft()
cnt += 1
for dz, dx, dy in dirs:
nz, nx, ny = cz+dz, cx+dx, cy+dy
if 0 <= nz < Z and 0 <= nx < X and 0 <= ny < Y:
if grid[nz][nx][ny] == '.' and comp[nz][nx][ny] == -1:
comp[nz][nx][ny] = cid
q.append((nz,nx,ny))
comp_size.append(cnt)
cid += 1
best = 0
seen = [0]*len(comp_size)
stamp = 1
for x in range(X):
for y in range(Y):
total = 0
stamp += 1
for z in range(Z):
if grid[z][x][y] == '.':
c = comp[z][x][y]
if seen[c] != stamp:
seen[c] = stamp
total += comp_size[c]
best = max(best, total)
print(best)
return solve()
# custom tests
assert run("""1
1 1 1
.
""") == "1", "minimum case"
assert run("""1
2 1 2
.
#
.
#
""") == "2", "vertical split same column"
assert run("""1
2 2 1
..
..
""") == "4", "single layer full gas"
| Test input | Expected output | What it validates |
|---|---|---|
| 1x1x1 gas | 1 | minimal single component |
| vertical alternating | 2 | column aggregation over multiple depths |
| full plane gas | 4 | one giant component across layer |
Edge Cases
One important edge case is when gas cells are isolated singletons. In this case, each gas cell forms its own component, and the answer for a column is simply the number of gas cells in that column. The BFS phase assigns a unique component id per cell, so no merging happens incorrectly, and the column scan adds each singleton exactly once.
Another edge case is a single huge component spanning the entire grid. Here, every gas cell receives the same component id. During column scanning, the first encountered gas cell triggers inclusion of the entire component, and all other cells are ignored due to the timestamp check. This prevents overcounting even though many cells in the same column belong to the same reservoir.
A final subtle case is multiple disconnected components stacked vertically in the same column. The scan correctly accumulates them because each component id is distinct, and the per-column visited mechanism ensures each contributes exactly once even if it appears in multiple z-levels.