/// <summary>
/// Solves a partial problem.
/// </summary>
/// <param name="partProblem">DVRP partial problem to solve.</param>
/// <param name="timeout">Timeout for computations.</param>
/// <returns>Solution of the given problem.</returns>
public DvrpSolution Solve(DvrpPartialProblem partProblem, TimeSpan timeout)
{
var resultCarsRoutes = new List <int> [_dvrpProblem.VehicleCount];
for (var i = 0; i < _dvrpProblem.VehicleCount; i++)
{
resultCarsRoutes[i] = new List <int>();
}
var min = partProblem.ApproximateResult;
var n = _dvrpProblem.Requests.Length;
var max = new int[n];
max[0] = -1;
var part = partProblem.SetBegin;
_timer.Reset();
_timer.Start();
// let's check all the partitions
for (ulong i = 0; IsLowerSet(part, partProblem.SetEnd); ++i)
{
if (_timer.Elapsed.TotalSeconds >= timeout.TotalSeconds)
{
State = UCCTaskSolver.TaskSolver.TaskSolverState.Timeout;
break;
}
for (var j = 1; j < n; ++j)
{
if (max[j - 1] < part[j - 1])
{
max[j] = part[j - 1];
}
else
{
max[j] = max[j - 1];
}
}
var p = n - 1;
while (part[p] == max[p] + 1)
{
part[p] = 0;
p = p - 1;
}
part[p] = part[p] + 1;
var breaking = false;
var cap = new List <int>();
var listOfRoutes = new List <List <int> >();
for (var j = 0; j < part.Length; ++j)
{
if (listOfRoutes.Count == part[j])
{
// With this version of algorthm we assume that one car can do only one ride
if (listOfRoutes.Count == _dvrpProblem.VehicleCount)
{
breaking = true;
break;
}
listOfRoutes.Add(new List <int>
{
j
});
cap.Add(_dvrpProblem.Requests[j].Demand);
}
else
{
listOfRoutes[part[j]].Add(j);
cap[part[j]] += (_dvrpProblem.Requests[j].Demand);
// The road has to be served in one ride
if (cap[part[j]] >= -_dvrpProblem.VehicleCapacity)
{
continue;
}
breaking = true;
var tmp = Math.Max(max[j], part[j]);
for (var k = j + 1; k < part.Length; k++)
{
part[k] = ++tmp;
}
break;
}
}
// if any of partitions doesn't fulfil the requirements
if (breaking)
{
continue;
}
double distance = 0;
var carsRoutes = new List <int> [_dvrpProblem.VehicleCount];
for (var j = 0; j < listOfRoutes.Count; ++j)
{
distance += _tspSolver.Solve(listOfRoutes[j], min, out carsRoutes[j]);
if (distance < min)
{
continue;
}
int k = 0;
for (int l = 0; l <= j; l++)
{
k = Math.Max(listOfRoutes[l][listOfRoutes[l].Count - 1], k);
}
var tmp = Math.Max(max[k], part[k]);
for (k += 1; k < part.Length; k++)
{
part[k] = ++tmp;
}
break;
}
if (min <= distance)
{
continue;
}
min = distance;
for (var j = 0; j < _dvrpProblem.VehicleCount; j++)
{
if (carsRoutes[j] == null)
{
resultCarsRoutes[j] = new List <int>();
}
else
{
resultCarsRoutes[j] = new List <int>(carsRoutes[j]);
}
}
}
var routes = new int[_dvrpProblem.VehicleCount][];
for (var i = 0; i < routes.Length; i++)
{
routes[i] = resultCarsRoutes[i].ToArray();
}
_timer.Stop();
return(new DvrpSolution(min, routes));
}
/// <summary>
/// Solves a partial problem.
/// </summary>
/// <param name="partProblem">DVRP partial problem to solve.</param>
/// <param name="timeout">Timeout for computations.</param>
/// <returns>Solution of the given problem.</returns>
public DvrpSolution Solve(DvrpPartialProblem partProblem, TimeSpan timeout)
{
var resultCarsRoutes = new List<int>[_dvrpProblem.VehicleCount];
for (var i = 0; i < _dvrpProblem.VehicleCount; i++)
{
resultCarsRoutes[i] = new List<int>();
}
var min = partProblem.ApproximateResult;
var n = _dvrpProblem.Requests.Length;
var max = new int[n];
max[0] = -1;
var part = partProblem.SetBegin;
_timer.Reset();
_timer.Start();
// let's check all the partitions
for (ulong i = 0; IsLowerSet(part, partProblem.SetEnd); ++i)
{
if (_timer.Elapsed.TotalSeconds >= timeout.TotalSeconds)
{
State = UCCTaskSolver.TaskSolver.TaskSolverState.Timeout;
break;
}
for (var j = 1; j < n; ++j)
{
if (max[j - 1] < part[j - 1])
max[j] = part[j - 1];
else
max[j] = max[j - 1];
}
var p = n - 1;
while (part[p] == max[p] + 1)
{
part[p] = 0;
p = p - 1;
}
part[p] = part[p] + 1;
var breaking = false;
var cap = new List<int>();
var listOfRoutes = new List<List<int>>();
for (var j = 0; j < part.Length; ++j)
{
if (listOfRoutes.Count == part[j])
{
// With this version of algorthm we assume that one car can do only one ride
if (listOfRoutes.Count == _dvrpProblem.VehicleCount)
{
breaking = true;
break;
}
listOfRoutes.Add(new List<int>
{
j
});
cap.Add(_dvrpProblem.Requests[j].Demand);
}
else
{
listOfRoutes[part[j]].Add(j);
cap[part[j]] += (_dvrpProblem.Requests[j].Demand);
// The road has to be served in one ride
if (cap[part[j]] >= -_dvrpProblem.VehicleCapacity) continue;
breaking = true;
var tmp = Math.Max(max[j], part[j]);
for (var k = j + 1; k < part.Length; k++)
{
part[k] = ++tmp;
}
break;
}
}
// if any of partitions doesn't fulfil the requirements
if (breaking)
continue;
double distance = 0;
var carsRoutes = new List<int>[_dvrpProblem.VehicleCount];
for (var j = 0; j < listOfRoutes.Count; ++j)
{
distance += _tspSolver.Solve(listOfRoutes[j], min, out carsRoutes[j]);
if (distance < min) continue;
int k = 0;
for (int l = 0; l <= j; l++)
{
k = Math.Max(listOfRoutes[l][listOfRoutes[l].Count - 1], k);
}
var tmp = Math.Max(max[k], part[k]);
for (k += 1; k < part.Length; k++)
{
part[k] = ++tmp;
}
break;
}
if (min <= distance) continue;
min = distance;
for (var j = 0; j < _dvrpProblem.VehicleCount; j++)
{
if (carsRoutes[j] == null)
resultCarsRoutes[j] = new List<int>();
else
resultCarsRoutes[j] = new List<int>(carsRoutes[j]);
}
}
var routes = new int[_dvrpProblem.VehicleCount][];
for (var i = 0; i < routes.Length; i++)
{
routes[i] = resultCarsRoutes[i].ToArray();
}
_timer.Stop();
return new DvrpSolution(min, routes);
}
