107 lines
3 KiB
Python
107 lines
3 KiB
Python
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#!/usr/bin/env python
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import sys
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from dataclasses import dataclass
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from math import ceil
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from typing import Dict, List
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@dataclass
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class Ingredient:
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name: str
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quantity: int
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@dataclass
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class ReactionEquation:
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quantity: int
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inputs: List[Ingredient]
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Reactions = Dict[str, ReactionEquation]
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def solve_for(n: int, reactions: Reactions) -> int:
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ore_needed = 0
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wanted = [("FUEL", n)]
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excess: Dict[str, int] = {}
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def provide_ingredient(name: str, wanted_quantity: int) -> None:
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nonlocal ore_needed
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nonlocal excess
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nonlocal wanted
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if name == "ORE":
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ore_needed += wanted_quantity # There's no recipy for this one
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return
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if name in excess:
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# Take from excess
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if excess[name] > wanted_quantity:
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excess[name] -= wanted_quantity
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return # Nothing left to do
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wanted_quantity -= excess[name]
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del excess[name] # Took everything
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if wanted_quantity == 0: # In case we provided just enough by excess
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return
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equation = reactions[name]
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reaction_num = ceil(wanted_quantity / equation.quantity)
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for ingredient in equation.inputs:
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needed_quantity = ingredient.quantity * reaction_num
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provide_ingredient(ingredient.name, needed_quantity)
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produced_quantity = equation.quantity * reaction_num
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excess_quantity = produced_quantity - wanted_quantity
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if excess_quantity > 0:
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if name in excess:
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excess[name] += excess_quantity
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else:
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excess[name] = excess_quantity
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while len(wanted) != 0:
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provide_ingredient(*(wanted.pop()))
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return ore_needed
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def main() -> None:
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reactions: Reactions = {}
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def parse_react(l: str) -> None:
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def parse_ingredient(i: str) -> Ingredient:
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quantity, name = i.strip().split(" ")
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return Ingredient(name, int(quantity))
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input_list, output_str = l.split("=>")
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inputs = [i for i in map(parse_ingredient, input_list.split(", "))]
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output = parse_ingredient(output_str)
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reactions[output.name] = ReactionEquation(output.quantity, inputs)
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for line in sys.stdin.readlines():
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parse_react(line)
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for_one = solve_for(1, reactions)
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target = 1000000000000
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# Educated guesses about minimum and maximum fuel needed to use 'target' ORE
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min_fuel_needed = target // for_one
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max_fuel_needed = 2 * min_fuel_needed
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while min_fuel_needed < max_fuel_needed:
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# We already know that minimum value is valid, offset it by one for the search
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mid = ceil((min_fuel_needed + max_fuel_needed) / 2)
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mid_res = solve_for(mid, reactions)
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if mid_res > target:
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# Exclude the maximum that was already searched
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max_fuel_needed = mid - 1
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else:
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# Keep the valid minimum value
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min_fuel_needed = mid
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print(max_fuel_needed)
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if __name__ == "__main__":
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main()
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