我知道这是一个旧帖子,但我有同样的问题,我终于找到了答案。
查看文档,我猜想interval_map的Subtractability和重叠功能的聚合保证了减法作为删除操作工作。
在将区间值对插入到区间映射中导致插入的区间值对与区间值对发生冲突的情况下,将加法或减法传播到区间映射的关联值是一个非常有成果的概念,即已经在 interval_map 中。这种操作传播称为重叠聚合。
假设我必须将 unix 时间戳的间隔与一些记录 id(整数)相匹配。根据这个答案,我想出了这个 MWE:
// interval_map_mwe.cpp
#include <map>
#include <set>
#include <climits>
#include <boost/icl/interval.hpp>
#include <boost/icl/interval_map.hpp>
// Set of IDs that cover a time interval
typedef std::set<unsigned int> IDSet_t;
// interval tree from intervals of timestamps to a set of ids
typedef boost::icl::interval_map<time_t, IDSet_t> IMap_t;
// a time interval
typedef boost::icl::interval<time_t> Interval_t;
#include <iostream>
// from https://stackoverflow.com/a/22027957
inline std::ostream& operator<< (std::ostream& S, const IDSet_t& X)
{
S << '(';
for (IDSet_t::const_iterator it = X.begin(); it != X.end(); ++it) {
if (it != X.begin()) {
S << ',';
}
S << *it;
}
S << ')';
return S;
}
int main(int argc, const char *argv[])
{
(void)argc; // suppress warning
(void)argv; // suppress warning
IMap_t m;
IDSet_t s;
s.insert(1);
s.insert(2);
m += std::make_pair(Interval_t::right_open(100, 200), s);
s = IDSet_t();
s.insert(3);
s.insert(4);
m += std::make_pair(Interval_t::right_open(200, 300), s);
s = IDSet_t();
s.insert(5);
s.insert(6);
m += std::make_pair(Interval_t::right_open(150, 250), s);
std::cout << "Initial map: " << std::endl;
std::cout << m << std::endl;
// find operation
IMap_t::const_iterator it = m.find(175);
std::cout << "Interval that covers 175: ";
std::cout << it->first << std::endl;
std::cout << "Ids in interval: " << it->second << std::endl;
// partially remove 5 from interval (160,180)
s = IDSet_t();
s.insert(5);
m -= std::make_pair(Interval_t::right_open(160, 180), s);
std::cout << "map with 5 partially removed:" << std::endl;
std::cout << m << std::endl;
// completelly remove 6
s = IDSet_t();
s.insert(6);
// Note: maybe the range of the interval could be shorter if you can somehow obtain the minimum and maximum times
m -= std::make_pair(Interval_t::right_open(0, UINT_MAX), s);
std::cout << "map without 6: " << std::endl;
std::cout << m << std::endl;
// remove a time interval
m -= Interval_t::right_open(160, 170);
std::cout << "map with time span removed: " << std::endl;
std::cout << m << std::endl;
return 0;
}
使用 g++ 4.4.7 编译:
g++ -Wall -Wextra -std=c++98 -I /usr/include/boost148/ interval_map_mwe.cpp
我得到的输出是
Initial map:
{([100,150)->{1 2 })([150,200)->{1 2 5 6 })([200,250)->{3 4 5 6 })([250,300)->{3 4 })}
Interval that covers 175: [150,200)
Ids in interval: (1,2,5,6)
map with 5 partially removed:
{([100,150)->{1 2 })([150,160)->{1 2 5 6 })([160,180)->{1 2 6 })([180,200)->{1 2 5 6 })([200,250)->{3 4 5 6 })([250,300)->{3 4 })}
map without 6:
{([100,150)->{1 2 })([150,160)->{1 2 5 })([160,180)->{1 2 })([180,200)->{1 2 5 })([200,250)->{3 4 5 })([250,300)->{3 4 })}
map with time span removed:
{([100,150)->{1 2 })([150,160)->{1 2 5 })([170,180)->{1 2 })([180,200)->{1 2 5 })([200,250)->{3 4 5 })([250,300)->{3 4 })}
注意: MWE 中的数字可以认为是随机的。我发现用小数字来推理这个例子更容易。