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November 6, 2019 03:12
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require 'priority_queue' | |
def solve(n, max_discs, l, t, m) | |
dp = Array.new(n + 1) { a = Array.new(n + 1, Float::INFINITY); a[0] = 0; a } | |
cut = Array.new(n + 1) { Array.new(n + 1) } | |
cumulate_ts = t.inject([0]) { |result, tvalue| result << result.last + tvalue } | |
(1..max_discs).each do |i| | |
k = 1 | |
pq = PriorityQueue.new | |
(1..n).each do |j| | |
cost = dp[i - 1][j - 1] | |
cost += 1 if j > 1 && m[j - 2] == m[j - 1] | |
pq.push [cost, j], cost * n + (n - j) # cost が異なるなら cost が小さい方を、cost が等しいなら index が大きい方を返す | |
while cumulate_ts[j] - cumulate_ts[k - 1] > l | |
k += 1 | |
end | |
while pq.min_key[1] < k | |
pq.delete_min | |
end | |
min = pq.min_key | |
if dp[i][j] > min[0] | |
dp[i][j] = min[0] | |
cut[i][j] = min[1] | |
end | |
end | |
if dp[i][n] < Float::INFINITY # 解を見つけた | |
partitions = [] | |
k = n | |
i.downto(1).each do |j| | |
partitions << [cut[j][k]-1..k-1] # cut の値や k は1始まりであることに注意 | |
k = cut[j][k] - 1 | |
end | |
return [dp[i][n], partitions.reverse] | |
end | |
end | |
nil | |
end | |
n = 5; max_discs = 5; l = 30 | |
t = [10, 10, 10, 10, 10] | |
m = [1, 1, 2, 2, 2] | |
cost, sep = solve(n, max_discs, l, t, m) | |
p "cost: #{cost}, separation: #{sep}" |
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