I have trouble printing out elements in set.
std::set<triple, Compare> edges;
for (int i = 0; i < n; i++)
for (std::list<std::pair<int, int>>::iterator j = graph[i].begin(); j != graph[i].end(); j++)
edges.insert(makeTriple((*j).second, i, (*j).first));
for (std::set<triple, Compare>::iterator j = edges.begin(); j != edges.end(); j++)
printf("%d and %d\n\n", (*j).first + 1, (*j).second + 1);
Only 7 elements are printed of of 13. Compare function looks like:
bool operator()(const triple &a, const triple &b) const
{
if (a.distance == b.distance && a.first == b.first)
return (a.second < b.second);
if (a.distance == b.distance && a.second == b.second)
return (a.first < b.first);
return (a.distance < b.distance);
}
Your Compare function does not fulfill the requirement
The return value of the function call operation applied to an object of a type satisfying
Compare, when contextually converted tobool, yieldstrueif the first argument of the call appears before the second in the strict weak ordering relation induced by this type, andfalseotherwise.
The most obvious fix would be:
bool operator()(const triple &a, const triple &b) const {
if (a.distance == b.distance) {
if(a.first == b.first) return a.second < b.second;
else return a.first < b.first;
} else return a.distance < b.distance;
}
This can however be done simpler with std::tie from <tuple>:
bool operator()(const triple &a, const triple &b) const {
return
std::tie(a.distance, a.first, a.second) < std::tie(b.distance, b.first, b.second);
}
Collected from the Internet
Please contact [email protected] to delete if infringement.
Comments