Question

我正在担任儿童游戏的足球比赛经理。可能会有很多回合，每个回合中的每个团队都会互相对抗。给定每个回合的组合，我需要比赛日历，并且同时进行比赛的团队数量最多。有没有一种方法可以对组合数组进行排序以达到结果？

我试图“制作”某种算法，但找不到最佳算法。在提供的代码中，保证两个团队可以同时玩，但不能再玩了。

teams = [1,2,3,4,5,6,7,8] # number of teams is variable

pairings = teams.combination(2).to_a
pairings.shuffle!
calendar = []
calendar << pairings.slice!(0)

while pairings.any?
    p = calendar.slice(-1)
    a = p[0]
    b = p[1]
    matched = false
    pairings.each do |pairing|
        next if pairing.include? a
        next if pairing.include? b
        matched = true
        calendar << pairings.delete(pairing)
        break
    end
    unless matched
        p1 = pairings.slice!(0)
        prev = false
        calendar.each do |pairing|
            if pairing.include?(a) || pairing.include?(b)
                prev = true
                next
            end
            unless prev
                i = calendar.index(pairing)
                calendar.insert(i, p1)
                break
            end
            prev = false
        end
    end
end

Answer 1

谷歌搜索条件提示我来了这个解决方案，非常感谢！

teams = [1,2,3,4,5]

teams.push "x" if teams.length.odd?

first = [teams.slice!(0)]

teams.length.times do
    half = Integer(teams.length / 2)
    row1 = first + teams.slice(0...half)
    row2 = teams.slice(half...teams.length).reverse
    for i in 0..half do
        next if row1[i] == "x"
        next if row2[i] == "x"
        puts "#{row1[i]} - #{row2[i]}"
    end
    teams.unshift(teams.pop)
end

给定数组的组合将它们排列在不同的数组中，这样就不会出现双精度

1 个答案: