這是一個簡單的動態規劃問題。
設總長度爲N，任務長度爲L。
讓F(T)是能夠從子間隔(T, N)被選擇的任務的最大數量，則每單位時間T處，有3種可能性：
- 沒有開始於T.
任務
- 有一項任務從T開始，但我們不會將其包含在結果集中。
- 有一個任務從T開始，我們將它包含在結果集中。
案例1很簡單，我們只有F(T) = F(T + 1)。
在2/3的情況下，注意選擇是開始T意味着我們必須拒絕接受這個任務運行時，即T和T + L之間的啓動所有任務的任務。所以我們得到F(T) = max(F(T + 1), F(T + L) + 1)。
最後，F(N) = 0。所以你只需從F(N)開始，然後回到F(0)。
編輯：這將給你最大間隔數，但不是滿足你的2個約束的集合。您對約束條件的解釋不清楚，所以我不確定如何在那裏幫助您。特別是，我不能說出什麼是約束條件，因爲你的示例集合的所有解決方案顯然是相同的。
編輯2：所要求的一些進一步的解釋：
考慮您發佈的例子，我們有N = 11和L = 2。有些任務開始於T = 0, 3, 4, 5, 6, 9。從F(11) = 0開始和向後工作：
F(11) = 0F(10) = F(11) = 0（由於沒有任務開始T = 10）F(9) = max(F(10), F(11) + 1) = 1- ...
最終我們得到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|
編輯3：好吧，我很好奇這件事，所以我寫了一個解決方案，所以不妨將它發佈。這將爲您提供具有最多任務，最少差距和最小最小差距的集合。在討論的例子輸出是：
(0, 2) -> (4, 6) -> (6, 8) -> (9, 11)(0, 2) -> (4, 6) -> (8, 10)
很顯然，我做出的正確性任何保證！ :)
私人課程任務 public int Start {get;組; } public int Length {get;組; } 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。但你的嘗試在哪裏？ –
一個很好的問題，但我認爲它不適合SO。 –
你是否試圖靠某種方式來成功？ 夥計，這不是歐拉項目。 – Vitaliy