Python
Just putting the solution to part 3 here, because these solutions are pretty long. Took me a while to figure out that an optimal return path from the bottom of the volcano burn can bleed into the left quadrant.
from collections import deque
import heapq as hq
import re
DIRECTIONS_CARDINAL = [(-1, 0), (0, 1), (1, 0), (0, -1)]
DIRECTIONS_ALL = [
*DIRECTIONS_CARDINAL, # cardinal
(-1, -1), (-1, 1), (1, -1), (1, 1) # diagonal
]
# yields all 8 possible neighbors
# lava spreads in all 8 directions
def yield_all_neighbors(i, j, m, n):
for di, dj in DIRECTIONS_ALL:
ni, nj = i + di, j + dj
if 0 <= ni < m and 0 <= nj < n:
yield ni, nj
# yields only the 4 possible cardinal neighbors
# the dragonduck can only move in cardinal directions
def yield_cardinal_neighbors(i, j, m, n):
for di, dj in DIRECTIONS_CARDINAL:
ni, nj = i + di, j + dj
if 0 <= ni < m and 0 <= nj < n:
yield ni, nj
# finds a special character in the grid
def find_char(data: str, ch: str):
i = 0
for row in data.splitlines():
j = row.find(ch)
if j != -1:
return (i, j)
i += 1
assert False, f"Character {ch} not found in grid"
# returns squared distance between two points
def get_dist_sq(a, b):
x1, y1 = a
x2, y2 = b
return (x2 - x1) ** 2 + (y2 - y1) ** 2
# Generator to burn cells permanently one step at a time
def bfs_burn(grid, volcano):
m, n = len(grid), len(grid[0])
queue = deque([volcano])
visited = set()
step = 0
while queue:
nq = len(queue)
for _ in range(nq):
i, j = queue.popleft()
if (i, j) in visited:
continue
# skip cells outside the current step radius
# but do not mark them as visited yet
if get_dist_sq((i, j), volcano) > (step * step):
continue
visited.add((i, j))
# burn cell permanently
grid[i][j] = -1
for ni, nj in yield_all_neighbors(i, j, m, n):
if (ni, nj) in visited:
continue
queue.append((ni, nj))
step += 1
# yield here to allow enclosing the volcano after each step
yield step
def part3(data: str):
# initialize variables
start = find_char(data, 'S')
volcano = find_char(data, '@')
data = re.sub(r'[@S]', '0', data)
grid = [[int(c) for c in row] for row in data.splitlines()]
m, n = len(grid), len(grid[0])
# initialize lava generator
lava_spreader = bfs_burn(grid, volcano)
# Finds shortest path from start to dest within time limit
# Dijkstra's algorithm with constraints
def find_shortest_path(start: tuple[int, int], dest: tuple[int, int], volcano_col: int, time_limit, map_half):
candidates = [(0, start[0], start[1])]
visited = set()
while candidates:
cost, i, j = hq.heappop(candidates)
if (i, j) in visited:
continue
visited.add((i, j))
if (i, j) == dest:
# if we are processing the destination, we have found the shortest path to it
return cost
if cost > time_limit:
# if we have exceeded the time limit, prune this path
continue
# remember: the dragonduck can only move in cardinal directions
for ni, nj in yield_cardinal_neighbors(i, j, m, n):
# skip visited or burned cells
if (ni, nj) in visited or grid[ni][nj] == -1:
continue
# if processing left half, skip paths going to the right of the volcano
if map_half == 'left' and nj > volcano_col:
continue
# if processing right half, skip paths going to the left of the volcano (ONLY for bottom quadrant)
# allow moving into the left half in the top quadrant (!!!)
elif map_half == 'right' and (ni > m // 2 and nj < volcano_col):
continue
# add new candidate path
hq.heappush(candidates, (cost + grid[ni][nj], ni, nj))
# we couldn't reach the destination within time limit
# fail by returning the maximum possible cost
return 9 * m * n
# try increasing lava spreads
while True:
# spread lava one more step
step = next(lava_spreader)
# calculate time limit
time = step * 30
# our paths from the start must connect at the bottom of the lava burn
burn_bottom_boundary = (volcano[0] + step, volcano[1])
time_taken = find_shortest_path(start, burn_bottom_boundary, volcano[1], time, 'left')
if time_taken >= time:
# if we already exceeded time, quit early
continue
time_taken += find_shortest_path(burn_bottom_boundary, start, volcano[1], time - time_taken, 'right')
if time_taken >= time:
# if we exceeded time, try next lava spread
continue
# we successfully enclosed the volcano within time
burn_radius = step - 1
return time_taken * burn_radius