哪个 STL 容器用于存储具有基于密钥的访问的有序数据？
假设我有一个 Person 对象的集合，每个对象如下所示：
class Person
{
string Name;
string UniqueID;
}
现在，这些对象必须存储在一个容器中，该容器允许我对它们进行排序，以便我可以给项目 X 轻松找到项目 X+1 和 X-1 。
但是，我还需要基于 UniqueID 的快速访问，因为集合很大，线性搜索无法解决。
我当前的“解决方案”是将 std::list 与 std::map 结合使用。该列表保存 Persons（用于有序访问），并且映射用于将 UniqueID 映射到对列表项的引用。更新“容器”通常涉及更新地图和列表。
它有效，但我觉得应该有一种更聪明的方法来做到这一点，也许是 boost:bimap。建议？
编辑：我对“订购”的要求有些混乱。解释一下，对象是从文件中按顺序流入的，并且容器中项目的“顺序”应与文件顺序匹配。该顺序与 ID 无关。
Let's say I have a collection of Person objects, each of which looks like this:
class Person
{
string Name;
string UniqueID;
}
Now, the objects must be stored in a container which allows me to order them so that I can given item X easily locate item X+1 and X-1.
However, I also need fast access based on the UniqueID, as the collection will be large and a linear search won't cut it.
My current 'solution' is to use a std::list in conjunction with a std::map. The list holds the Persons (for ordered access) and the map is used to map UniqueID to a reference to the list item. Updating the 'container' typically involves updating both map and list.
It works, but I feel there should be a smarter way of doing it, maybe boost:bimap. Suggestions?
EDIT: There's some confusion about my requirement for "ordering". To explain, the objects are streamed in sequentially from a file, and the 'order' of items in the container should match the file order. The order is unrelated to the IDs.
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boost:bimap是最明显的选择。bimap基于boost::multi_index，但bimap简化了语法。就我个人而言，我更喜欢boost::multi_index而不是boost::bimap，因为它允许将来轻松地向Person结构添加更多索引。boost:bimapis the most obvious choice.bimapis based onboost::multi_index, butbimaphas simplified syntax. Personally I will preferboost::multi_indexoverboost::bimapbecause it will allow to easily add more indices to thePersonstructure in the future.没有标准库容器可以满足您的需求 - 因此您必须使用两个容器或 Boost 解决方案。如果使用两个容器，在几乎所有情况下，我通常更喜欢向量或双端队列而不是列表。
There is no Standard Library container that does what you want - so you will have to use two containers or the Boost solution. If using two containers, I would normally prefer a vector or a deque over a list, in almost all circumstances.
为什么不使用两张地图，一张以 Person 作为键，另一张以 UniqueId 作为键，但这需要更新它们。
您可以创建一个回调函数，每当有任何变化时都会更新两个地图。
Why not to use two maps , one having Person as Key and another one having UniqueId as Key, but that requires updating both of them.
you can create a callback function which updates both the maps whenever there is any change.