I don't have a framework to profile the operations but I'd certainly change the code to reuse the readily allocated vector. In addition, I'd hoist the initial intersection out of the loop. Also, std::back_inserter()
should make sure that elements are added in the correct location rather than in the beginning:
int func()
{
vector<vector<unsigned> > t = some_initialization();
if (t.empty()) {
return;
}
vector<unsigned> intersectedValues(t[0]);
vector<unsigned> tempIntersectedSubjects;
for (std::vector<std::vector<unsigned>>::size_type i(1u);
i < t.size() && !intersectedValues.empty(); ++i) {
std::set_intersection(t[i].begin(), t[i].end(),
intersectedValues.begin(), intersectedValues.end(),
std::back_inserter(tempIntersectedSubjects);
std::swap(intersectedValues, tempIntersectedSubjects);
tempIntersectedSubjects.clear();
}
}
I think this code has a fair chance to be faster. It may also be reasonable to intersect the sets different: instead of keeping one set and intersecting with that you could create a new intersection for pairs of adjacent sets and then intersect the first sets with their respect adjacent ones:
std::vector<std::vector<unsigned>> intersections(
std::vector<std::vector<unsigned>> const& t) {
std::vector<std::vector<unsigned>> r;
std::vector<std::vector<unsignned>>::size_type i(0);
for (; i + 1 < t.size(); i += 2) {
r.push_back(intersect(t[i], t[i + 1]));
}
if (i < t.size()) {
r.push_back(t[i]);
}
return r;
}
std::vector<unsigned> func(std::vector<std::vector<unsigned>> const& t) {
if (t.empty()) { /* deal with t being empty... */ }
std::vector<std::vector<unsigned>> r(intersections(t))
return r.size() == 1? r[0]: func(r);
}
Of course, you wouldn't really implement it like this: you'd use Stepanov's binary counter to keep the intermediate sets. This approach assumes that the result is most likely non-empty. If the expectation is that the result will be empty that may not be an improvement.
与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…