Anders Englöf Ytterström
7c2b4c835a
Part 1 was easily solved by Dijkstras (using heapq). Tricky part was to find the way to serialize the queue items for the heappush and heappop to behave as required. Part 2 was incredible hard to figure out, and I could not do it by myself. From the subreddit, it is hinted that a traditional set of visited nodes to determine algorithm exit would not work. After some laboration and readin some hints, I managed to find a solution where the code keeps tracks using a defaultdict.
78 lines
2.1 KiB
Python
78 lines
2.1 KiB
Python
from heapq import heappop, heappush
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from output import D, matrix
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def solve(data):
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grid, H, W = matrix(data)
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S = [(r, c) for r in range(H) for c in range(W) if grid[r][c] == "S"][0]
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E = [(r, c) for r in range(H) for c in range(W) if grid[r][c] == "E"][0]
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Q = [(0, S, 1)]
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p1 = float("inf")
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SE = dict()
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ES = dict()
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seen = set()
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while Q:
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cost, pos, dir = heappop(Q)
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r, c = pos
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if grid[r][c] == "#":
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continue
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if (r, c, dir) in seen:
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continue
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seen.add((r, c, dir))
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if (r, c, dir) not in SE:
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SE[(r, c, dir)] = cost
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if pos == E:
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p1 = min(p1, cost)
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continue
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for inc, delta, facing in [
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(1, D[dir], dir),
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(1000, (0, 0), (dir + 1) % 4),
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(1000, (0, 0), (dir - 1) % 4),
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]:
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nc = cost + inc
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dr, dc = delta
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heappush(Q, (nc, (r + dr, c + dc), facing))
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Q = [(0, E, 0), (0, E, 1), (0, E, 2), (0, E, 3)]
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seen = set()
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while Q:
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cost, pos, dir = heappop(Q)
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r, c = pos
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if grid[r][c] == "#":
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continue
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if (r, c, dir) in seen:
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continue
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seen.add((r, c, dir))
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if (r, c, (dir + 2) % 4) not in ES:
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ES[(r, c, (dir + 2) % 4)] = cost
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if pos == S:
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continue
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for inc, delta, facing in [
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(1, D[dir], dir),
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(1000, (0, 0), (dir + 1) % 4),
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(1000, (0, 0), (dir - 1) % 4),
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]:
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nc = cost + inc
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dr, dc = delta
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heappush(Q, (nc, (r + dr, c + dc), facing))
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p2 = set()
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for r in range(H):
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for c in range(W):
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for d in range(4):
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if (
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(r, c, d) in SE
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and (r, c, d) in ES
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and SE[(r, c, d)] + ES[(r, c, d)] == p1
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):
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p2.add((r, c))
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return p1, len(p2)
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if __name__ == "__main__":
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with open("./input/16.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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