Efficient way to enumerate unique undirected paths - c++

Question

Given a subset of nodes {1,2,...,N} is there any STL or boost function that returns unique undirected tours over all of them?
std::next_permutation() gives all N! directed tours, where 1-2-...-N is different from N-N-1-...-2-1.
However, in this case, I don't want both of them, but only one of them. Essentially, I would like to enumerate only N! / 2 of the tours.
The following code that uses std::next_permutation() and unordered_set works, but is there anything more efficient? The following code essentially generates all N! directed tours and discards half of them after checking against an unordered_set().
#include <vector>
#include <unordered_set>
#include <algorithm>
#include <boost/functional/hash.hpp>
template <typename T, typename U> bool unorderedset_val_there_already_add_if_not(std::unordered_set<T, U>& uos, T& val) {
if (uos.find(val) != uos.end())
return true;//val already there
uos.insert(val);
return false;//Value is new.
}
int main() {
std::vector<int> sequence{ 1, 2, 3};
std::unordered_set<std::vector<int>, boost::hash<std::vector<int>>> uos;
do {
printf("Considering ");
for (std::size_t i = 0; i < sequence.size(); i++)
printf("%d ", sequence[i]);
printf("\n");
std::vector<int> rev_sequence = sequence;
std::reverse(rev_sequence.begin(), rev_sequence.end());
if (unorderedset_val_there_already_add_if_not(uos, sequence) || unorderedset_val_there_already_add_if_not(uos, rev_sequence)) {
printf("Already there by itself or its reverse.\n");
}
else {
printf("Sequence and its reverse are new.\n");
}
} while (std::next_permutation(sequence.begin(), sequence.end()));
getchar();
}
That is, given {1,2,3}, I only want to enumerate (1-2-3), (1-3-2) and (2-1-3). The other three permutations (2-3-1), (3-1-2) and (3-2-1) should not be enumerated because their reverse sequence have already been enumerated.

Answer 1

If you want to stay with next_permutation rather than make own generator routine, the simplest way is filter out a half of permutation with some condition.
Very simple one: the last element should be larger than the first one.
#include <vector>
#include <algorithm>
#include "stdio.h"
int main() {
std::vector<int> sequence{ 1, 2, 3, 4};
do {
if (sequence[sequence.size()-1] > sequence[0]) {
for (std::size_t i = 0; i < sequence.size(); i++)
printf("%d ", sequence[i]);
printf("\n");
}
} while (std::next_permutation(sequence.begin(), sequence.end()));
getchar();
}
1 2 3 4
1 2 4 3
1 3 2 4
1 3 4 2
1 4 2 3
1 4 3 2
2 1 3 4
2 1 4 3
2 3 1 4
2 4 1 3
3 1 2 4
3 2 1 4
Possible own implementation:
Generate all pairs (start; end) where start < end
Generate all permutations of `n-2` values without start and end
For every permutation make {start, permutation.., end}
1 ... 2 + permutations of {3, 4}
1 3 4 2
1 4 3 2
1 ... 3 + permutations of {2,4}
1 2 4 3
1 4 2 3
...
3 ... 4 + permutations of {1, 2}
3 1 2 4
3 2 1 4
...