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