Anders Englöf Ytterström
f5c3ee938a
Solution is still 100 lines, but now it is based on a singular BFS oriented box push algorithm.
115 lines
3 KiB
Python
115 lines
3 KiB
Python
from output import D, matrix
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def solve(data):
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O = {k: v for k, v in zip("^>v<", D)}
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grid, movements = data.split("\n\n")
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grid, H, W = matrix(grid)
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movements = "".join(movements.split())
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grid_1 = {
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(r, c): value for r, row in enumerate(grid) for c, value in enumerate(row)
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}
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grid_2 = {}
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for pos, value in grid_1.items():
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r2, c2 = pos
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match value:
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case "O":
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grid_2[(r2, c2 * 2)] = "["
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grid_2[(r2, c2 * 2 + 1)] = "]"
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case "@":
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grid_2[(r2, c2 * 2)] = "@"
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grid_2[(r2, c2 * 2 + 1)] = "."
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case _:
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grid_2[(r2, c2 * 2)] = value
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grid_2[(r2, c2 * 2 + 1)] = value
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answers = []
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for grid, W in [(grid_1, W), (grid_2, W * 2)]:
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pos = next((r, c) for r in range(H) for c in range(W) if grid[(r, c)] == "@")
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for move in movements:
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r, c = pos
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dr, dc = O[move]
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move_boxes = True
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blocked = False
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if (r + dr, c + dc) not in grid:
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continue
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match grid[(r + dr, c + dc)]:
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case ".":
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move_boxes = False
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case "#":
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move_boxes = False
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continue
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if move_boxes:
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seen = set()
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Q = [(r + dr, c + dc)]
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while Q:
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rr, cc = Q.pop(0)
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if (rr, cc) in seen:
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continue
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if (rr, cc) not in grid:
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continue
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if grid[(rr, cc)] == "#":
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blocked = True
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continue
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seen.add((rr, cc))
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if grid[(rr, cc)] in ".":
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continue
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Q.append((rr + dr, cc + dc))
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if dr != 0:
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match grid[(rr, cc)]:
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case "]":
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Q.append((rr, cc - 1))
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case "[":
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Q.append((rr, cc + 1))
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if blocked:
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continue
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patchset = {
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(rr, cc): (rr + dr, cc + dc)
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for rr, cc in seen
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if grid[(rr, cc)] != "."
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}
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dotset = set(seen) - set(patchset.values())
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for oldpos, newpos in sorted(patchset.items())[:: -(dr + dc)]:
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value = grid[oldpos]
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grid[newpos] = value
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for dot in dotset:
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grid[dot] = "."
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grid[(r, c)] = "."
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grid[(r + dr, c + dc)] = "@"
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pos = r + dr, c + dc
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answers.append(sum(k[0] * 100 + k[1] for k, v in grid.items() if v in "O["))
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p1, p2 = answers
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return p1, p2
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if __name__ == "__main__":
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with open("./input/15.txt", "r") as f:
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inp = f.read().strip()
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p1, p2 = solve(inp)
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print(p1)
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print(p2)
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