C# 是否有标准算法将重叠对象平衡到桶中？
我有一群用户，具有给定的开始和结束时间，例如：C# 是否有标准算法将重叠对象平衡到桶中？,c#,.net,algorithm,combinatorics,C#,.net,Algorithm,Combinatorics,我有一群用户，具有给定的开始和结束时间，例如： { Name = "Peter", StartTime = "10:30", EndTime = "11:00" }, { Name = "Dana", StartTime = "11:00", EndTime = "12:30" }, { Name = "Raymond", StartTime = "10:30", EndTime = "14:00" }, { Name = "Egon", StartTime = "12:00", EndTime
{ Name = "Peter", StartTime = "10:30", EndTime = "11:00" },
{ Name = "Dana", StartTime = "11:00", EndTime = "12:30" },
{ Name = "Raymond", StartTime = "10:30", EndTime = "14:00" },
{ Name = "Egon", StartTime = "12:00", EndTime = "13:00" },
{ Name = "Winston", StartTime = "10:00", EndTime = "12:00" }
（标记组合数学，因为我认为这是它适用的数学领域，如果错了请纠正我）基于你的问题，我可能会做一些事情，比如首先创建一个名为“Person”pr的类。为该类指定“名称”、“开始时间”和“结束时间”属性 然后把它们放在Person类型的有序列表中。（另外，我目前将时间存储为双倍。只需将时间的任何小数部分乘以60/100，就可以将它们转换回时间 接下来，您将创建一个存储桶列表，如果需要，您可以向其中添加新的存储桶。然后，您将根据您定义的阈值对列表进行排序，如果正在比较的两个对象根据该阈值重叠，则这两个对象都将进入该存储桶。如果它们不重叠，则移动到下一个存储桶，如果存在重叠，则将其添加到下一个存储桶直到你到达最后一个桶为止。如果你已经完成了所有的桶并且仍然没有重叠，那么为该对象创建一个新的桶
class MainFunc
{
static void Main(string[] args)
{
//makes a function to check if 2 values overlap over a threshold
//st stands for start time and et stands for end time
public bool IsWithinThreshold(double st1, double st2, double et1, double et2)
{
double threshold = .5;
if(st1 >= et2 || st2 >= et1)
{
return false
}
else
{
if(st1+threshold <= et2 && st1+threshold <= et1 || st2+threshold <= et1 && st2+threshold <=et2)
{
return true;
}
else
{
return false;
}
}
}
// makes objects of type Person with attributes of name, start time, and end time
Person Peter = new Person();
Peter.name = "Peter"
Peter.start_time = 10.5
Peter.end_time = 11.0
Person Dana = new Person();
Dana.name = "Dana"
Peter.start_time = 11.0
Peter.end_time = 12.5
Person Raymond = new Person();
Raymond.name = "Raymond"
Raymond.start_time = 10.5
Raymond.end_time = 14.0
Person Egon = new Person();
Egon.name = "Egon"
Egon.start_time = 12.0
Egon.end_time = 13.0
Person Winston = new Person();
Winston.name = "Winston"
Winston.start_time = 10.0
Winston.end_time = 12.0
//puts objects of type Person into an unordered list
List<Person> people = new List<Person>();
people.Add(Peter);
people.Add(Dana);
people.Add(Raymond);
people.Add(Egon);
people.Add(Winston);
//sets up a list of lists of People (Buckets in our case)
List<List<Person>> Buckets = new List<List<Person>>;
//sets up an intial Bucket and adds the first person on the list to it
List<Person> Bucketinitial = new List<Person>;
Bucketinitial.add(people[0]);
for(var i = 1; i < people.Count; i++)
{
for(var j = 0; j< Buckets.count; j++)
{
//sets a checker to make sure that all objects in a given Bucket overlap with the person we are checking
bool overlap = true;
for(var k = 0; k< Buckets[k].count; k++)
{
overlap = overlap & IsWithinThreshold(people[i].start_time,Buckets[j][k].start_time,people[i].end_time,Buckets[j][k].end_time)
}
if (overlap == true)
{
Buckets[j].add(people[i])
}
//if all the objects in a bucket don't overlap with the person...
//... make a new Bucket with that person
else
{
List<Person> NewBucket = new List<Person>;
NewBucket.add(people[i]);
Buckets.add(NewBucket);
}
}
}
}
}
class MainFunc
{
静态void Main（字符串[]参数）
{
//生成一个函数，用于检查两个值是否在阈值上重叠
//st代表开始时间，et代表结束时间
公共布尔值在阈值内（双st1、双st2、双et1、双et2）
{
双阈值=.5；
如果（st1>=et2 | | st2>=et1）
{
返回错误
}
其他的
{
if（st1+阈值tl；dr:win（O（排序（n））时间）的动态规划）
首先，请注意，按开始时间顺序连续扣合是可以的
建议（碎片整理）：让a、b、c、d
成为不同的用户，以便开始时间（a）≤ 开始时间（b）≤ 开始时间（c）≤ 开始时间（d）
。如果X
和Y
是有效的存储桶，则a、c∈ X
和b，d∈ Y
，然后X-{c}∪ {b}
和Y-{a}∪ {d}
也是有效的存储桶
我只知道如何通过冗长的案例分析来证明这一点（略）
结果是，你可以假装把一个段落分成几行，其中“段落”是按开始时间顺序排列的用户列表，每个“行”是一个桶。根据Knuth和Plass，有一种优化换行的算法，其中给定行的惩罚或多或少是一个任意函数。例如，您可以使4个桶的成本为0，3个桶的成本为1，2个桶的成本为2，1个桶的成本为100。您可以修改算法，将区间树合并到加快搜索速度
按开始时间排序
将项目添加到间隔树
创建一个空桶
从列表中选择第一项
使用间隔树的间隔搜索，查找第一个装满存储桶的项目的阈值时间内的最早项目
从列表中删除带扣的项目
如果列表为空，则停止，否则转至步骤4
基本上，您是以间隔步骤（由您可配置的阈值给定）从左向右移动，在移动过程中使用间隔树快速查询最近的项目。是否希望仅根据30分钟（阈值）来订购存储桶重叠或基于谁重叠最多？@或避免任何重叠都可以。我们可以调整最小重叠（例如，最小60分钟），但重叠60分钟的人和重叠180分钟的人一样好。我认为间隔树会很好地工作。我做了一个间隔树程序。它肯定没有算法那么干净，但基本上是我下面的答案。*注意，我的间隔可能会因此而变短，所以当你把代码放在你最喜欢的com中时piler，请确保对其进行响应，这样您就有了完美的标记行代码。
class MainFunc
{
static void Main(string[] args)
{
//makes a function to check if 2 values overlap over a threshold
//st stands for start time and et stands for end time
public bool IsWithinThreshold(double st1, double st2, double et1, double et2)
{
double threshold = .5;
if(st1 >= et2 || st2 >= et1)
{
return false
}
else
{
if(st1+threshold <= et2 && st1+threshold <= et1 || st2+threshold <= et1 && st2+threshold <=et2)
{
return true;
}
else
{
return false;
}
}
}
// makes objects of type Person with attributes of name, start time, and end time
Person Peter = new Person();
Peter.name = "Peter"
Peter.start_time = 10.5
Peter.end_time = 11.0
Person Dana = new Person();
Dana.name = "Dana"
Peter.start_time = 11.0
Peter.end_time = 12.5
Person Raymond = new Person();
Raymond.name = "Raymond"
Raymond.start_time = 10.5
Raymond.end_time = 14.0
Person Egon = new Person();
Egon.name = "Egon"
Egon.start_time = 12.0
Egon.end_time = 13.0
Person Winston = new Person();
Winston.name = "Winston"
Winston.start_time = 10.0
Winston.end_time = 12.0
//puts objects of type Person into an unordered list
List<Person> people = new List<Person>();
people.Add(Peter);
people.Add(Dana);
people.Add(Raymond);
people.Add(Egon);
people.Add(Winston);
//sets up a list of lists of People (Buckets in our case)
List<List<Person>> Buckets = new List<List<Person>>;
//sets up an intial Bucket and adds the first person on the list to it
List<Person> Bucketinitial = new List<Person>;
Bucketinitial.add(people[0]);
for(var i = 1; i < people.Count; i++)
{
for(var j = 0; j< Buckets.count; j++)
{
//sets a checker to make sure that all objects in a given Bucket overlap with the person we are checking
bool overlap = true;
for(var k = 0; k< Buckets[k].count; k++)
{
overlap = overlap & IsWithinThreshold(people[i].start_time,Buckets[j][k].start_time,people[i].end_time,Buckets[j][k].end_time)
}
if (overlap == true)
{
Buckets[j].add(people[i])
}
//if all the objects in a bucket don't overlap with the person...
//... make a new Bucket with that person
else
{
List<Person> NewBucket = new List<Person>;
NewBucket.add(people[i]);
Buckets.add(NewBucket);
}
}
}
}
}