Anders Englöf Ytterström
8683885438
For part 1, BFS is used since there is no need to visit each splitter more than once to count visited splitters. For part 2, the code initially removed the visited check. This failed miserably, losing momentum aound Y=54-56. A recursive function with memoization solves it much faster.
50 lines
1 KiB
Python
50 lines
1 KiB
Python
import functools
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from output import grid
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def solve(data):
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G = grid(data, o="^")
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p1 = set()
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p2 = 0
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H = len(data.split())
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W = len(data.split()[0])
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S = (0, data.split()[0].index("S"))
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Q = [S]
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while Q:
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y, x = Q.pop(0)
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if y == H:
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continue
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if (y, x) in p1:
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continue
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if (y, x) in G:
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Q.append((y, x - 1))
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Q.append((y, x + 1))
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p1.add((y, x))
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else:
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Q.append((y + 1, x))
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p1 = len(p1)
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@functools.cache
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def _timelines(p):
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y, x = p
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if not 0 <= y < H or not 0 <= x < W:
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return 1
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if p in G:
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return _timelines((y, x - 1)) + _timelines((y, x + 1))
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return _timelines((y + 1, x))
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p2 = _timelines(S)
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return p1, p2
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if __name__ == "__main__":
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with open("./input/07.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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assert p1 == 1539
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assert p2 == 6479180385864
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