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。
设F(T)为子区间 可以选择的最大任务数(T, N)，则在每个单位时间 T，有 3 种可能性：
- 没有从 T 开始的任务。
- 有一个任务从 T 开始，但我们没有将它包含在结果集中。
- 有一个从 T 开始的任务，我们确实将它包含在结果集中。
案例 1 很简单，我们只有F(T) = F(T + 1).
在案例 2/3 中，请注意选择启动 a 的任务T意味着我们必须拒绝在此任务运行时启动的所有任务，即介于T和之间T + L。所以我们得到F(T) = max(F(T + 1), F(T + L) + 1).
最后，F(N) = 0. F(N)因此，您只需从F(0).
编辑：这将为您提供最大数量的间隔，但不是满足您的 2 个约束的集合。我不清楚您对约束的解释，所以我不确定如何帮助您。特别是，我无法说出约束 1 的含义，因为您的示例集的所有解决方案显然都是相等的。
编辑2：根据要求进行一些进一步的解释：
考虑您发布的示例，我们有N = 11和L = 2。有些任务从 开始T = 0, 3, 4, 5, 6, 9。从开始F(11) = 0和向后工作：
F(11) = 0F(10) = F(11) = 0（因为没有任务从 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，我对此很好奇，所以我写了一个解决方案，所以不妨发布它。这将为您提供任务最多、间隔最少和最小间隔最小的集合。问题中示例的输出是：
(0, 2) -> (4, 6) -> (6, 8) -> (9, 11)(0, 2) -> (4, 6) -> (8, 10)
显然，我不保证正确性！:)
私有类任务 { 公共 int 开始 { 获取；放; } 公共 int 长度 { 获取；放; } public int End { 获取 { 返回开始 + 长度；} }
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();
}