c# – 在电影院对朋友进行分组的算法

我认为你需要递归地执行它,但是你需要确保不要一遍又一遍地对同一组进行分区.这将为您提供指数执行时间.在我的解决方案中,它看起来我有O(n * n)(您可以为我验证它),请参阅下面的结果.另一件事是你提到的desirablity功能.我不知道这样的函数怎么样,但你可以比较2个分区.例如分区1 1 2 4比1 2 2 3更不可取,因为它有两个“1”.一般规则可能是“如果分区的人数与另一个分区相同,则分区不太可取”.有道理,坐在一起的人越多越好.我的解决方案采用这种方法来比较2个可能的分组,并得到你想要实现的结果.让我先向您展示一些结果,然后是代码.

var sut = new BrainTeaser();

        for (int n = 1; n <= 6; n++) {
            StringBuilder sb = new StringBuilder();
            sb.AppendFormat("{0} person{1}: ",n,n > 1 ? "s" : "");

            var array = sut.Solve(n).Select(x => x.ToString()).ToArray();
            sb.AppendLine(string.Join(",",array));

            Console.WriteLine(sb.ToString());
        }

1人：1

2人：2,11

3人：3,1 1,2 1 1

4人：4,1 3,1 1 2,1 1 1 1

5人：5,2 3,1 4,1 2 2,1 1 3,1 1 1 2,1 1 1 1 1

6人：6,1 5 1,1 1 4,1 1 2 2,1 1 1 3,1 1 1 1 2,1 1 1 1 1 1

性能看起来是O(n * n)：

var sut = new BrainTeaser();

 for (int n = 1; n <= 40; n++) {
   Stopwatch watch = new Stopwatch();
   watch.Start();
   var count = sut.Solve(n).Count();
   watch.Stop();
   Console.WriteLine("Problem solved for {0} friends in {1} ms. Number of solutions {2}",watch.ElapsedMilliseconds,count);
}

Problem solved for 1 friends in 17 ms. Number of solutions 1
Problem solved for 2 friends in 49 ms. Number of solutions 2
Problem solved for 3 friends in 2 ms. Number of solutions 3
Problem solved for 4 friends in 1 ms. Number of solutions 5
Problem solved for 5 friends in 0 ms. Number of solutions 7
Problem solved for 6 friends in 2 ms. Number of solutions 11
Problem solved for 7 friends in 0 ms. Number of solutions 15
Problem solved for 8 friends in 0 ms. Number of solutions 22
Problem solved for 9 friends in 1 ms. Number of solutions 30
Problem solved for 10 friends in 1 ms. Number of solutions 42
Problem solved for 11 friends in 4 ms. Number of solutions 56
Problem solved for 12 friends in 4 ms. Number of solutions 77
Problem solved for 13 friends in 7 ms. Number of solutions 101
Problem solved for 14 friends in 9 ms. Number of solutions 135
Problem solved for 15 friends in 15 ms. Number of solutions 176
Problem solved for 16 friends in 21 ms. Number of solutions 231
Problem solved for 17 friends in 30 ms. Number of solutions 297
Problem solved for 18 friends in 43 ms. Number of solutions 385
Problem solved for 19 friends in 61 ms. Number of solutions 490
Problem solved for 20 friends in 85 ms. Number of solutions 627
Problem solved for 21 friends in 117 ms. Number of solutions 792
Problem solved for 22 friends in 164 ms. Number of solutions 1002
Problem solved for 23 friends in 219 ms. Number of solutions 1255
Problem solved for 24 friends in 300 ms. Number of solutions 1575
Problem solved for 25 friends in 386 ms. Number of solutions 1958
Problem solved for 26 friends in 519 ms. Number of solutions 2436
Problem solved for 27 friends in 677 ms. Number of solutions 3010
Problem solved for 28 friends in 895 ms. Number of solutions 3718
Problem solved for 29 friends in 1168 ms. Number of solutions 4565
Problem solved for 30 friends in 1545 ms. Number of solutions 5604
Problem solved for 31 friends in 2025 ms. Number of solutions 6842
Problem solved for 32 friends in 2577 ms. Number of solutions 8349
Problem solved for 33 friends in 3227 ms. Number of solutions 10143
Problem solved for 34 friends in 4137 ms. Number of solutions 12310
Problem solved for 35 friends in 5300 ms. Number of solutions 14883
Problem solved for 36 friends in 6429 ms. Number of solutions 17977
Problem solved for 37 friends in 8190 ms. Number of solutions 21637
Problem solved for 38 friends in 10162 ms. Number of solutions 26015
Problem solved for 39 friends in 12643 ms. Number of solutions 31185

现在让我发布解决方案中涉及的3个类：

public class BrainTeaser {
    /// <summary>
    /// The possible groupings are returned in order of the 'most' desirable first. Equivalent groupings are not returned (e.g. 2 + 1 vs. 1 + 2). Only one representant
    /// of each grouping is returned (ordered ascending. e.g. 1 + 1 + 2 + 4 + 5)
    /// </summary>
    /// <param name="numberOfFriends"></param>
    /// <returns></returns>
    public IEnumerable<PossibleGrouping> Solve(int numberOfFriends) {
        if (numberOfFriends == 1) {
            yield return new PossibleGrouping(1);
            yield break;
        }
        HashSet<PossibleGrouping> possibleGroupings = new HashSet<PossibleGrouping>(new PossibleGroupingComparer());
        foreach (var grouping in Solve(numberOfFriends - 1)) {
            // for each group we create 'n+1' new groups 
            // 1 + 1 + 2 + 3 + 4 
            // Becomes
            //      (1+1) + 1 + 2 + 3 + 4  we can add a friend to the first group
            //      1 + (1+1) + 2 + 3 + 4  we can add a friend to the second group
            //      1 + 1 + (2+1) + 3 + 4  we can add a friend to the third group
            //      1 + 1 + 2 + (3+1) + 4  we can add a friend to the forth group
            //      1 + 1 + 2 + 3 + (4+1) we can add a friend to the fifth group
            //      (1 + 1 + 2 + 3 + 4) + 1  friend has to sit alone

            AddAllPartitions(grouping,possibleGroupings);
        }
        foreach (var possibleGrouping in possibleGroupings.OrderByDescending(x => x)) {
            yield return possibleGrouping;
        }
    }

    private void AddAllPartitions(PossibleGrouping grouping,HashSet<PossibleGrouping> possibleGroupings) {
        for (int i = 0; i < grouping.FriendsInGroup.Length; i++) {
            int[] newFriendsInGroup = (int[]) grouping.FriendsInGroup.Clone();
            newFriendsInGroup[i] = newFriendsInGroup[i] + 1;
            possibleGroupings.Add(new PossibleGrouping(newFriendsInGroup));
        }
        var friendsInGroupWithOneAtTheEnd = grouping.FriendsInGroup.Concat(new[] {1}).ToArray();
        possibleGroupings.Add(new PossibleGrouping(friendsInGroupWithOneAtTheEnd));
    }
}

/// <summary>
/// A possible grouping of friends. E.g.
/// 1 + 1 + 2 + 2 + 4 (10 friends). The array is sorted by the least friends in an group.
/// </summary>
public class PossibleGrouping : IComparable<PossibleGrouping> {
    private readonly int[] friendsInGroup;

    public int[] FriendsInGroup {
        get { return friendsInGroup; }
    }

    private readonly int sum;

    public PossibleGrouping(params int[] friendsInGroup) {
        this.friendsInGroup = friendsInGroup.OrderBy(x => x).ToArray();
        sum = friendsInGroup.Sum();
    }

    public int Sum {
        get { return sum; }
    }

    /// <summary>
    /// determine which group is more desirable. Example:
    /// Consider g1: 1 + 2 + 3 + 4 vs g2: 1 + 1 + 2 + 2 + 4  
    /// Group each sequence by the number of occurrences:
    /// 
    /// group   | g1   | g2
    /// --------|-------------
    /// 1       | 1    | 2
    /// ----------------------
    /// 2       | 1    | 2
    /// ----------------------
    /// 3       | 1    | 0
    /// ----------------------
    /// 4       | 1    | 1
    /// ----------------------
    /// 
    /// Sequence 'g1' should score 'higher' because it has 'less' 'ones' (least desirable) elements. 
    /// 
    /// If both sequence would have same number of 'ones',we'd compare the 'twos'.
    /// 
    /// </summary>
    /// <param name="other"></param>
    /// <returns></returns>
    public int CompareTo(PossibleGrouping other) {
        var thisGroup = (from n in friendsInGroup group n by n).ToDictionary(x => x.Key,x => x.Count());

        var otherGroup = (from n in other.friendsInGroup group n by n).ToDictionary(x => x.Key,x => x.Count());

        return WhichGroupIsBetter(thisGroup,otherGroup);
    }

    private int WhichGroupIsBetter(IDictionary<int,int> thisGroup,IDictionary<int,int> otherGroup) {
        int maxNumberOfFriendsInAGroups = Math.Max(thisGroup.Keys.Max(),otherGroup.Keys.Max());

        for (int numberOfFriendsInGroup = 1;
             numberOfFriendsInGroup <= maxNumberOfFriendsInAGroups;
             numberOfFriendsInGroup++) {
            // zero means that the current grouping does not contain a such group with 'numberOfFriendsInGroup'
            // in the example above,e.g. group '3'
            int thisNumberOfGroups = thisGroup.ContainsKey(numberOfFriendsInGroup)
                                         ? thisGroup[numberOfFriendsInGroup]
                                         : 0;
            int otherNumberOfGroups = otherGroup.ContainsKey(numberOfFriendsInGroup)
                                          ? otherGroup[numberOfFriendsInGroup]
                                          : 0;

            int compare = thisNumberOfGroups.CompareTo(otherNumberOfGroups);

            if (compare != 0) {
                // positive score means that the other group has more occurrences. e.g. 'this' group might have 2 groups with each 2 friends,// but the other solution might have 3 groups with each 2 friends. It's obvious that (because both solutions must sum up to the same value)
                // this 'solution' must contain a grouping with more than 3 friends which is more desirable.
                return -compare;
            }
        }
        // they must be 'equal' in this case.
        return 0;
    }

    public override string ToString() {
        return string.Join("+",friendsInGroup.Select(x => x.ToString()).ToArray());
    }
}

public class PossibleGroupingComparer : EqualityComparer<PossibleGrouping> {
    public override bool Equals(PossibleGrouping x,PossibleGrouping y) {
        return x.FriendsInGroup.SequenceEqual(y.FriendsInGroup);
    }

    /// <summary>
    /// may not be the best hashcode function. for alternatives look here: http://burtleburtle.net/bob/hash/doobs.html
    /// I got this code from here: https://stackoverflow.com/questions/3404715/c-sharp-hashcode-for-array-of-ints
    /// </summary>
    /// <param name="obj"></param>
    /// <returns></returns>
    public override int GetHashCode(PossibleGrouping obj) {
        var array = obj.FriendsInGroup;

        int hc = obj.FriendsInGroup.Length;
        for (int i = 0; i < array.Length; ++i) {
            hc = unchecked(hc*314159 + array[i]);
        }
        return hc;
    }
}

现在解决方案：

brainteaser类执行递归.这个类中的一个技巧是在hashset中使用自定义比较器(PossibleGroupingComparer).这将确保当我们计算新分组时(例如1 1 2 vs 2 1 1),这些将被视为相同(我们的集合将仅包含每个等效分组的一个代表).这应该将指数运行时间减少到O(n ^ 2).

下一个技巧是可以对结果进行排序,因为我们的PossibleGroupings类实现了IComparable. Compare()方法的实现使用了上面提到的想法.此方法基本上包含此解决方案中的salt,如果您希望以不同方式对其进行分组,则应仅修改此方法.

我希望你能理解代码,否则让我知道.我试图让它可读,并不关心性能.例如,您可以在将它们返回给调用者之前对这些分组进行排序,递归中的排序不会带来太大的影响.

但有一条评论：一个典型的场景可能是电影院已经“已经”预订了许多座位而且不允许“任何”分区.在这里,您需要获取所有分区,然后逐个检查是否可以用于当前的电影院.这工作,但成本不必要的CPU.相反,我们可以使用输入来减少递归次数并缩短总体执行时间.也许有人想为此发布一个解决方案;)