Here is the greedy algorithm – always the best place to start.
allocate all teams
IF not all sections covered
output -1
stop
mark all teams non-critical
flag_improved = true
WHILE( flag_improved == true )
flag_improved = false
find most expensive section
find most expensive non-critical team on most expensive section
IF team found that can be removed without leaving a section uncovered
remove team
flag_improved = true
ELSE
mark team critical
output cost - sum coverage of remaining teams
Here is the main function of a C++ application implementing this algorithm
main()
{
std::vector<cTeam> teams;
int bridge_section_count;
Input( bridge_section_count, teams);
// allocate all teams
for (int s = 0; s < bridge_section_count; s++)
Bridge.insert(std::pair(s, std::vector<cTeam>(0)));
for (auto &t : teams)
for (int s = t.myRange.first; s <= t.myRange.second; s++)
{
Bridge[s].push_back(t);
}
// check every section has at least one team allocated
if (!IsCovered())
{
std::cout << "-1\n";
exit(1);
}
// loop while improvements are being found
bool flag_improved = true;
while (flag_improved)
{
flag_improved = false;
auto most_expensive_section = find_most_expensive_section();
while (1)
{
// loop over teams allocated to most expensive section
std::vector<cTeam>::iterator most_expensive_team;
if (!find_most_expensive_non_critical_team(
most_expensive_team,
most_expensive_section))
{
break;
}
// check can team be removed without leaving section uncovered
if (testRemoval(*most_expensive_team))
{
// remove team
doRemoval(*most_expensive_team);
flag_improved = true;
break;
}
else
{
// this team is critical, it cannot be removed
most_expensive_team->myCritical = true;
}
}
}
printResult();
}
The complete application code is at https://gist.github.com/JamesBremner/ada6210a8517671abd45e882c97d526d
solved How to cover a range using a set of ranges with minimal overlap?