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:
将来可能会增加更多选项:
背景如下:
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。
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 Decreasing和Tabu 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。