53 lines
2.0 KiB
Kotlin
53 lines
2.0 KiB
Kotlin
import java.io.File
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import java.io.InputStream
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import kotlin.math.max
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fun main(args: Array<String>) {
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val numWorkers = 5
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val inputStream = File("input").inputStream()
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val inputLines = inputStream.bufferedReader().use { it.readText() }.split("\n")
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var map: HashMap<Char, HashSet<Char>> = HashMap()
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for (line in inputLines) {
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val step = line[36]
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val dep = line[5]
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// Make sure values that are only referenced as dependencies
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// (i. e. the first step in our chain) are also added to the set.
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map.getOrPut(dep, { HashSet<Char>() })
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val currentDeps = map.getOrPut(step, { HashSet<Char>() })
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currentDeps.add(dep)
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}
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var workers = MutableList<Pair<Char, Int>>(numWorkers, { Pair( ' ', 0 ) })
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var timer = 0
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// While there are still objects to be processed or at least one worker is still working
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while (map.size > 0 || true in workers.map{ w -> w.second > 0 }) {
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// Only remove the element from dependencies once it’s been worked on
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for (worker in workers) {
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if (worker.second == 0) {
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map.values.map{ v -> v.remove(worker.first) }
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}
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}
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while (map.size > 0 && 0 in workers.map{ w -> w.second }) {
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// The keyset in Kotlin is already sorted, so we can just take the first value
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val candidates = map.entries.map{ e -> Pair(e.key, e.value.size) }.filter{ e -> e.second == 0 }.map{ e -> e.first }
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if (candidates.size == 0) {
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break
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}
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val next = candidates[0]
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val currentWorker = workers.indexOf(workers.find{ w -> w.second == 0 })
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// The char 'A' is 65, but we want 60 + its position in the alphabet
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workers[currentWorker] = Pair(next, next.toInt() - 4)
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map.remove(next)
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}
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timer++
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workers = workers.map{ w -> Pair(w.first, max(0, w.second - 1)) }.toMutableList()
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}
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println(timer)
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}
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