调度最佳记录列表的算法

时间:2012-09-03 08:59:36

标签: performance algorithm language-agnostic scheduling

我正在寻找一种算法,用于为给定以下项目列表的记录设备生成最佳记录列表:

Image link here http://img692.imageshack.us/img692/7952/recordlist.png

目前约束是:

  • 不存在任何重叠。
  • 在计算速度和已解决的冲突之间妥协。

将来可能会增加更多选项:

  • 用户可以拥有多个录制设备。
  • 用户可以为他/她的收藏计划建立录制优先级(1个最高 - 3个最低)。

背景如下:

  • 项目列表最长为1周。(当前时间 - 当前时间+ 1周)
  • 平均物品清单大小为100个项目,最多300个。

另外我想知道是否有办法生成最佳可能的记录列表(发送到记录的程序的百分比最高) 在这种情况下,无论处理时间如何,我们都无法解决100%的冲突。

提前致谢。

2 个答案:

答案 0 :(得分:1)

class Recording
{
    public int ProgrammeId { get; set; }
    public string ProgrammeTitle { get; set; }
    public DateTime StartTime { get; set; }
    public DateTime EndTime { get; set; }
    public int ChannelId { get; set; }
    public string ChannelName { get; set; }
    public int Weight { get { return 1; } } // Constant weight
}

贪婪的方法是按照start_time的增加顺序考虑这些程序。如果某个程序与之前选择的程序兼容,请选择它:

public static IEnumerable<Recording> GreedySelection(IList<Recording> data)
{
    data = data
        .OrderBy(r => r.StartTime)
        .ThenBy(r => r.EndTime)
        .ToList();

    DateTime? lastEnd = null;
    foreach (var rec in data)
    {
        if (lastEnd == null || rec.StartTime >= lastEnd.Value)
        {
            yield return rec;
            lastEnd = rec.EndTime;
        }
    }
}

要获得最佳加权解决方案,您可以使用动态编程:

public static IEnumerable<Recording> WeightedSelection(IList<Recording> data)
{
    data = data
        .OrderBy(r => r.EndTime)
        .ThenBy(r => r.StartTime)
        .ToList();

    int count = data.Count;
    var lastCompatible = new int?[count];

    // Compute the closest program before in time, that is compatible.
    // This nested loop might be optimizable in some way.
    for (int i = 0; i < count; i++)
    {
        for (int j = i - 1; j >= 0; j--)
        {
            if (data[j].EndTime <= data[i].StartTime)
            {
                lastCompatible[i] = j;
                break; // inner loop
            }
        }
    }

    // Dynamic programming to calculate the best path
    var optimalWeight = new int[count];
    var cameFrom = new int?[count];
    for (int i = 0; i < count; i++)
    {
        int weightWithItem = data[i].Weight;
        if (lastCompatible[i] != null)
        {
            weightWithItem += optimalWeight[lastCompatible[i].Value];
        }

        int weightWithoutItem = 0;
        if (i > 0) weightWithoutItem = optimalWeight[i-1];

        if (weightWithItem < weightWithoutItem)
        {
            optimalWeight[i] = weightWithoutItem;
            cameFrom[i] = i - 1;
        }
        else
        {
            optimalWeight[i] = weightWithItem;
            cameFrom[i] = lastCompatible[i];
        }
    }

    // This will give the programs in reverse order.
    for (int? i = count - 1; i != null; i = cameFrom[i.Value])
    {
        yield return data[i.Value];
    }
}

并非此版本包含重量,并尝试最大化重量总和。如果所有权重都设置为一(1),则两种算法的结果大小将具有相同的大小,因为大小等于权重。

贪婪的结果:

ProgrammeTitle           StartTime         EndTime
Star Trek                2012-09-03 02:05  2012-09-03 03:05
Everybody Loves Raymond  2012-09-03 06:00  2012-09-03 07:00
CSI                      2012-09-03 07:00  2012-09-03 08:00
Mythbusters              2012-09-03 08:00  2012-09-03 09:00
CSI                      2012-09-03 10:00  2012-09-03 11:00
Mythbusters              2012-09-03 11:00  2012-09-03 12:00
Star Trek                2012-09-04 02:05  2012-09-04 03:05
Everybody Loves Raymond  2012-09-04 06:00  2012-09-04 07:00
CSI                      2012-09-04 07:00  2012-09-04 08:00
Mythbusters              2012-09-04 08:00  2012-09-04 09:00
CSI                      2012-09-04 10:00  2012-09-04 11:00
Mythbusters              2012-09-04 11:00  2012-09-04 12:00

动态结果(已排序):

ProgrammeTitle           StartTime         EndTime
Everybody Loves Raymond  2012-09-03 03:00  2012-09-03 04:00
Everybody Loves Raymond  2012-09-03 06:00  2012-09-03 07:00
CSI                      2012-09-03 07:00  2012-09-03 08:00
CSI                      2012-09-03 08:30  2012-09-03 09:30
CSI                      2012-09-03 10:00  2012-09-03 11:00
Star Trek                2012-09-03 11:05  2012-09-03 12:05
Everybody Loves Raymond  2012-09-04 03:00  2012-09-04 04:00
Everybody Loves Raymond  2012-09-04 06:00  2012-09-04 07:00
CSI                      2012-09-04 07:00  2012-09-04 08:00
CSI                      2012-09-04 08:30  2012-09-04 09:30
CSI                      2012-09-04 10:00  2012-09-04 11:00
Star Trek                2012-09-04 11:05  2012-09-04 12:05

算法基于此文档:

答案 1 :(得分:0)

使用First Fit DecreasingTabu Search

此用例非常类似于护士排班(see this video):您不是将护班分配给护士,而是将项目分配给录像机,其中录像机是[recorder1,notRecorded]的列表。