这是一个简单的动态规划问题。
设总长度为N，任务长度为L。
让F(T)是能够从子间隔(T, N)被选择的任务的最大数量，则每单位时间T处，有3种可能性：
- 没有开始于T.
任务
- 有一项任务从T开始，但我们不会将其包含在结果集中。
- 有一个任务从T开始，我们将它包含在结果集中。
案例1很简单，我们只有F(T) = F(T + 1)。
在2/3的情况下，注意选择是开始T意味着我们必须拒绝接受这个任务运行时，即T和T + L之间的启动所有任务的任务。所以我们得到F(T) = max(F(T + 1), F(T + L) + 1)。
最后，F(N) = 0。所以你只需从F(N)开始，然后回到F(0)。
编辑：这将给你最大间隔数，但不是满足你的2个约束的集合。您对约束条件的解释不清楚，所以我不确定如何在那里帮助您。特别是，我不能说出什么是约束条件，因为你的示例集合的所有解决方案显然是相同的。
编辑2：所要求的一些进一步的解释：
考虑您发布的例子，我们有N = 11和L = 2。有些任务开始于T = 0, 3, 4, 5, 6, 9。从F(11) = 0开始和向后工作：
F(11) = 0F(10) = F(11) = 0（由于没有任务开始T = 10）F(9) = max(F(10), F(11) + 1) = 1- ...
最终我们得到F(0) = 4：
T |00|01|02|03|04|05|06|07|08|09|10|
F(T)| 4| 3| 3| 3| 3| 2| 2| 1| 1| 1| 0|
编辑3：好吧，我很好奇这件事，所以我写了一个解决方案，所以不妨将它发布。这将为您提供具有最多任务，最少差距和最小最小差距的集合。在讨论的例子输出是：
(0, 2) -> (4, 6) -> (6, 8) -> (9, 11)(0, 2) -> (4, 6) -> (8, 10)
很显然，我做出的正确性任何保证！ :)
私人课程任务 public int Start {get;组; } public int Length {get;组; } public int End {get {return Start + Length; }}
public override string ToString()
{
return string.Format("({0:d}, {1:d})", Start, End);
}
}
private class CacheEntry : IComparable
{
public int Tasks { get; set; }
public int Gaps { get; set; }
public int MinGap { get; set; }
public Task Task { get; set; }
public Task NextTask { get; set; }
public int CompareTo(object obj)
{
var other = obj as CacheEntry;
if (Tasks != other.Tasks)
return Tasks - other.Tasks; // More tasks is better
if (Gaps != other.Gaps)
return other.Gaps = Gaps; // Less gaps is better
return MinGap - other.MinGap; // Larger minimum gap is better
}
}
private static IList<Task> F(IList<Task> tasks)
{
var end = tasks.Max(x => x.End);
var tasksByTime = tasks.ToLookup(x => x.Start);
var cache = new List<CacheEntry>[end + 1];
cache[end] = new List<CacheEntry> { new CacheEntry { Tasks = 0, Gaps = 0, MinGap = end + 1 } };
for (int t = end - 1; t >= 0; t--)
{
if (!tasksByTime.Contains(t))
{
cache[t] = cache[t + 1];
continue;
}
foreach (var task in tasksByTime[t])
{
var oldCEs = cache[t + task.Length];
var firstOldCE = oldCEs.First();
var lastOldCE = oldCEs.Last();
var newCE = new CacheEntry
{
Tasks = firstOldCE.Tasks + 1,
Task = task,
Gaps = firstOldCE.Gaps,
MinGap = firstOldCE.MinGap
};
// If there is a task that starts at time T + L, then that will always
// be the best option for us, as it will have one less Gap than the others
if (firstOldCE.Task == null || firstOldCE.Task.Start == task.End)
{
newCE.NextTask = firstOldCE.Task;
}
// Otherwise we want the one that maximises MinGap.
else
{
var ce = oldCEs.OrderBy(x => Math.Min(x.Task.Start - newCE.Task.End, x.MinGap)).Last();
newCE.NextTask = ce.Task;
newCE.Gaps++;
newCE.MinGap = Math.Min(ce.MinGap, ce.Task.Start - task.End);
}
var toComp = cache[t] ?? cache[t + 1];
if (newCE.CompareTo(toComp.First()) < 0)
{
cache[t] = toComp;
}
else
{
var ceList = new List<CacheEntry> { newCE };
// We need to keep track of all subsolutions X that start on the interval [T, T+L] that
// have an equal number of tasks and gaps, but a possibly a smaller MinGap. This is
// because an earlier task may have an even smaller gap to this task.
int idx = newCE.Task.Start + 1;
while (idx < newCE.Task.End)
{
toComp = cache[idx];
if
(
newCE.Tasks == toComp.First().Tasks &&
newCE.Gaps == toComp.First().Gaps &&
newCE.MinGap >= toComp.First().MinGap
)
{
ceList.AddRange(toComp);
idx += toComp.First().Task.End;
}
else
idx++;
}
cache[t] = ceList;
}
}
}
var rv = new List<Task>();
var curr = cache[0].First();
while (true)
{
rv.Add(curr.Task);
if (curr.NextTask == null) break;
curr = cache[curr.NextTask.Start].First();
}
return rv;
}
public static void Main()
{
IList<Task> tasks, sol;
tasks = new List<Task>
{
new Task { Start = 0, Length = 2 },
new Task { Start = 3, Length = 2 },
new Task { Start = 4, Length = 2 },
new Task { Start = 5, Length = 2 },
new Task { Start = 6, Length = 2 },
new Task { Start = 9, Length = 2 },
};
sol = F(tasks);
foreach (var task in sol)
Console.Out.WriteLine(task);
Console.Out.WriteLine();
tasks = new List<Task>
{
new Task { Start = 0, Length = 2 },
new Task { Start = 3, Length = 2 },
new Task { Start = 4, Length = 2 },
new Task { Start = 8, Length = 2 },
};
sol = F(tasks);
foreach (var task in sol)
Console.Out.WriteLine(task);
Console.Out.WriteLine();
tasks = new List<Task>
{
new Task { Start = 0, Length = 5 },
new Task { Start = 6, Length = 5 },
new Task { Start = 7, Length = 3 },
new Task { Start = 8, Length = 9 },
new Task { Start = 19, Length = 1 },
};
sol = F(tasks);
foreach (var task in sol)
Console.Out.WriteLine(task);
Console.Out.WriteLine();
Console.In.ReadLine();
}
有趣。但你的尝试在哪里？ –
一个很好的问题，但我认为它不适合SO。 –
你是否试图靠某种方式来成功？ 伙计，这不是欧拉项目。 – Vitaliy