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- #!/usr/bin/env python3
- import sys
- import re
- from math import ceil
- def parse_blueprint(line):
- ore_ore, clay_ore, obs_ore, obs_clay, geo_ore, geo_obs = \
- map(int, re.findall(r'\d+', line)[1:])
- return (
- (0, geo_obs, 0, geo_ore),
- (0, 0, obs_clay, obs_ore),
- (0, 0, 0, clay_ore),
- (0, 0, 0, ore_ore),
- )
- def max_geo(blueprint, time):
- work = [(time, (0, 0, 0, 0), (0, 0, 0, 1))]
- best = 0
- # We can only make 1 robot per turn so we should only produce the maximum of
- # any single material cost per turn.
- max_robots = [time] + [max(c[ty] for c in blueprint) for ty in range(1, 4)]
- while work:
- time, materials, robots = state = work.pop()
- final_geo = materials[0] + time * robots[0]
- # Best case we'll keep building a geo robot each turn from now on.
- # If even that is not enough to improve, ignore the state.
- if final_geo + sum(range(1, time)) < best:
- continue
- # Accumulate geodes in the remaining time, and try to improve by
- # building more robots.
- best = max(best, final_geo)
- for ty, cost in enumerate(blueprint):
- # Skip time until we have accumulated the material cost, then wait 1
- # turn for production. A very low default denominator approximates
- # infinite build time for robots requiring unproduced materials.
- build_time = 1 + max(ceil(max(c - m, 0) / (r or 0.0001))
- for c, m, r in zip(cost, materials, robots))
- if build_time < time and robots[ty] < max_robots[ty]:
- newmat = tuple(m + r * build_time - c
- for c, m, r in zip(cost, materials, robots))
- newrobots = tuple(r + (i == ty) for i, r in enumerate(robots))
- work.append((time - build_time, newmat, newrobots))
- return best
- blueprints = list(map(parse_blueprint, sys.stdin))
- print(sum((i + 1) * max_geo(bp, 24) for i, bp in enumerate(blueprints)))
- a, b, c = (max_geo(bp, 32) for bp in blueprints[:3])
- print(a * b * c)
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