c# - 分发比赛“锦标赛友好”

Question

假设我有四支球队，ABCD，我想创建比赛，以便球队均匀地进行比赛：

不想要

• 甲对乙
• 甲与丙
• A与D
• 乙与丙
• B 与 D
• C与D

想要的

• 甲对乙
• C与D
• 乙与丙
• A与D
• 甲与丙
• B 与 D

换一种说法：我希望球队尽可能少地连续比赛。
目标语言是 C#，但我可以轻松翻译。

编辑：快速旁注，可以超过 4 个团队！

Answer 1

I think you may achieve what you need doing as follows. If you've got n number of teams, all the possible matches between teams can be represented with a Kn Complete graph.

The way I would come up with your desired sorting, is taking (removing) edges from that graph, one at a time, always trying to find an edge that matches teams that didn't match immediately before. Further more, I think this approach (with a little variation) can be used to generate the best way to maximize concurrent matches, if each time you take an edge you pick the one with the teams that haven't played in most time.

For simplicity, lets assume that teams are ints in the range of 0 to n-1. The graph can simply be represented with a boolean matrix. To pick the match with the teams that haven't played in most time, you can keep track of the last time each team played. For n teams, you'll have a total of n*(n-1)/2 number of matches.

IEnumerable<Tuple<int,int>> GenerateMatches(int n) {
    bool[,] pendingMatches = new bool[n,n];
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < i; j++)
            pendingMatches[i,j] = true;
    }

    int[] lastPlayed = new int[n];
    int totalMatches = n*(n-1)/2;
    for (int m = 1; m <= totalMatches; m++) {
        Tuple<int, int> t = null;
        int longestPlayed = -1;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < i; j++) {
                if (pendingMatches[i,j]) {
                    int moreRecentlyPlayed = Math.Max(lastPlayed[i], lastPlayed[j]);
                    int timeSinceLastPlay = m - moreRecentlyPlayed;
                    if (timeSinceLastPlay > longestPlayed) {
                        longestPlayed = timeSinceLastPlay;
                        t = Tuple.Create(i,j);
                    }
                }
            }
        }
        lastPlayed[t.Item1] = lastPlayed[t.Item2] = m;
        pendingMatches[t.Item1, t.Item2] = false;
        yield return t;
    }
}

The part that chooses the next match are the two most nested fors. The complexity of that part can be improved if you use some kind of priority queue adapted to adjust the priority of edges that involve the teams of the last picked edge after updating the lastPlayed array.

Hope that helps.

Answer 2

解决此问题的一种方法是通过以下步骤：

1. 创建一个包含匹配组合总数的集合。
2. 创建一个与步骤 1 中的集合长度相同的新集合。
3. 遍历步骤 1 中的项目，在步骤 2 中添加相同的项目，条件是下一个要添加的项目应该与上一个添加的项目之间的差异最大。

一些示例代码：

// Just a container to conveniently hold a match between two teams
public class Match
{
    public Match(string teamOne, string teamTwo)
    {
        TeamOne = teamOne;
        TeamTwo = teamTwo;
    }

    public string TeamOne { get; private set; }

    public string TeamTwo { get; private set; }

    public override string ToString()
    {
        return String.Format("{0} vs {1}", TeamOne, TeamTwo);
    }
}

public class MatchMaking
{
    public MatchMaking()
    {
        Teams = new List<string> { "A", "B", "C", "D", "E" };
    }

    public IList<string> Teams { get; private set; }

    public IList<Match> GetMatches()
    {
        var unorderedMatches = GetUnorderedMatches();

        // The list that we will eventually return
        var orderedMatches = new List<Match>();

        // Track the most recently added match
        Match lastAdded = null;

        // Loop through the unordered matches
        // Add items to the ordered matches
        // Add the one that is most different from the last added match
        for (var i = 0; i < unorderedMatches.Count; i++)
        {
            if (lastAdded == null)
            {
                lastAdded = unorderedMatches[i];
                orderedMatches.Add(unorderedMatches[i]);
                unorderedMatches[i] = null;
                continue;
            }

            var remainingMatches = unorderedMatches.Where(m => m != null);

            // Get the match which is the most different from the last added match
            // We do this by examining all of the unadded matches and getting the maximum difference
            Match mostDifferentFromLastAdded = null;
            int greatestDifference = 0;
            foreach (var match in remainingMatches)
            {
                var difference = GetDifference(lastAdded, match);
                if (difference > greatestDifference)
                {
                    greatestDifference = difference;
                    mostDifferentFromLastAdded = match;
                }

                if (difference == 2)
                {
                    break;
                }
            }

            // Add the most different match
            var index = unorderedMatches.ToList().IndexOf(mostDifferentFromLastAdded);
            lastAdded = unorderedMatches[index];
            orderedMatches.Add(unorderedMatches[index]);
            unorderedMatches[index] = null;
        }

        return orderedMatches;
    }

    // Create a list containing the total match combinations with an arbitrary order
    private List<Match> GetUnorderedMatches()
    {
        var totalNumberOfCombinations = AdditionFactorial(Teams.Count - 1);

        var unorderedMatches = new List<Match>(totalNumberOfCombinations);
        for (int i = 0; i < Teams.Count; i++)
        {
            for (int j = 0; j < Teams.Count; j++)
            {
                if (j <= i) continue;

                unorderedMatches.Add(new Match(Teams[i], Teams[j]));
            }
        }
        return unorderedMatches;
    }

    // Get the difference between two matches
    // 0 - no difference, 1 - only one team different, 2 - both teams different
    private int GetDifference(Match matchOne, Match matchTwo)
    {
        var matchOneTeams = new HashSet<string> { matchOne.TeamOne, matchOne.TeamTwo };
        var matchTwoTeams = new HashSet<string> { matchTwo.TeamOne, matchTwo.TeamTwo };
        var intersection = matchOneTeams.Intersect(matchTwoTeams);

        return (intersection.Count() - 2) * -1;
    }

    // Just a helper to get the total number of match combinations
    private int AdditionFactorial(int seed)
    {
        int result = 0;

        for (int i = seed; i > 0; i--)
        {
            result += i;
        }

        return result;
    }
}

public class Program
{
    private static void Main(string[] args)
    {
        var matchMaking = new MatchMaking();

        foreach (var match in matchMaking.GetMatches())
        {
            Console.WriteLine(match);
        }
    }
}