排序集没有严格的弱排序 [英] Sorted set without a strict weak ordering
问题描述
我有以下问题:考虑这个(简化的)结构:
struct任务{
int优先级;
std :: string说明;
//其他一些字段
};
现在,我想拥有一组所有任务并做一些工作。因此,我有一个相等运算符,它检查每个元素是否相等。
bool isEqual(const Task& lhs,const Task& rhs){
return lhs.priority == rhs.priority&&
lhs.description == rhs.description&&
//其他一些字段
;
}
为此,我使用了std :: unordered_set,效果很好。 / p>
但是现在我希望这些任务按集合中的优先级排序(以获得最高优先级的任务)。显然,这对于std :: unordered_set是不可能的,因此我尝试了使用以下较少运算符的std :: set:
bool lessTask(const Task& lhs,const Task& rhs){
return lhs.priority< rhs.priority;
}
但这意味着严格的弱排序,当两个任务相等时,优先级是相等的,这是我不希望的(我想维护我的isEqual方法进行相等性检查)。
什么是完成一组任务的最佳方法,我可以在其中真正快速地插入元素并且没有重复的条目(由isEqual函数定义),但是能够真正快速地检索具有最高优先级的任务吗?
我没有绑定到任何特定的STL容器,但不想使用任何第三方库(甚至都不使用Boost)。
首先写入 get_tie
:
// auto或decltype(auto)
auto get_tie(const Task& task){
return std :: tie(lhs.priority,lhs.description,/ *其他一些字段* /);
}
在C ++ 11中,您必须使用<$ c重复正文$ c>-> decltype 尾随返回类型,或使用宏来避免重复:
#define RETURNS(...)decltype(__ VA_ARGS__){return __VA_ARGS__; }
auto get_tie(const Task& task)->
RETURNS(std :: tie(lhs.priority,lhs.description,/ *其他一些字段* /))
一旦我们有一个简单的 get_tie
,您的问题就消失了。
bool isEqual(Task const& lhs,Task const& rhs){
return get_tie(lhs)== get_tie(rhs);
}
bool isLess(Task const& lhs,Task const& rhs){
return get_tie(lhs)< get_tie(rhs);
}
只需通过 isLess
到 std :: set
,或构建 std :: vector
和 std :: sort
使用 isLess
。
现在,如果您的比较在原始的 tie
引用,您可能必须用更复杂的东西替换 get_tie
。
I have the following problem: Consider this (simplified) structure:
struct Task {
int priority;
std::string description;
// Some other fields
};
Now I want to have a set of all tasks and do some work with it. Therefore I have an equality operator, which checks that every element is equal.
bool isEqual(const Task& lhs, const Task& rhs) {
return lhs.priority == rhs.priority &&
lhs.description == rhs.description &&
// Some other fields
;
}
For this I have used the std::unordered_set, which worked fine.
But now I want these tasks to be sorted by their priority(to get the highest priority task) in the set. Obviously this is not possible with an std::unordered_set, so I tried a std::set with the following less operator:
bool lessTask(const Task& lhs, const Task& rhs) {
return lhs.priority < rhs.priority;
}
But this implies by the strict weak ordering, that two tasks are equal, when the priority is equal, which I don't want(I want to maintain my isEqual method for equality checking).
What's the best way to accomplish a set of tasks, where I can insert elements really fast and have no duplicate entries(defined by my isEqual function), but are able to retrieve a task with the highest priority really fast?
I am not bound to any specific STL container, but doesn't want to use any third party libraries(not even boost).
First write get_tie
:
// auto or decltype(auto)
auto get_tie(const Task& task) {
return std::tie(lhs.priority, lhs.description, /* some other fields */ );
}
in C++11 you have to repeat the body with a ->decltype
trailing return type, or use a macro to avoid the repetition:
#define RETURNS(...) decltype(__VA_ARGS__) { return __VA_ARGS__; }
auto get_tie(const Task& task)->
RETURNS( std::tie(lhs.priority, lhs.description, /* some other fields */ ) )
once we have an easy get_tie
, your problem evaporates.
bool isEqual( Task const& lhs, Task const& rhs ) {
return get_tie(lhs)==get_tie(rhs);
}
bool isLess( Task const& lhs, Task const& rhs ) {
return get_tie(lhs) < get_tie(rhs);
}
simply pass isLess
to std::set
, or build a std::vector
and std::sort
it using isLess
.
Now, if your comparison doesn't really work on a raw tie
of references, you may have to replace get_tie
with something more complex.
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