void TabuSearch(){
int debug;
int number_of_iterations = 100;
int TabuLength = n;
get_first_path();
vector<vector<int> > tabuList(TabuLength, std::vector<int>(TabuLength));
vector <int> bestSol((int)current_path.size());
bestSol = current_path;
int shortest_path = GetTotalDistance(current_path);
for (int i = 0; i < number_of_iterations;i++) {
current_path = getBestNeighbour(tabuList, current_path);
int currCost = GetTotalDistance(current_path);
if (currCost < shortest_path) {
bestSol = current_path;
shortest_path = currCost;
}
}
printf("shortest_path = %d", shortest_path);
getchar();
getchar();
}
Data MatrixGraph::tabuSearch(uint tabuListSize, uint iterations)
{
clock_t overallTime = clock();
initRand();
if (!iterations)
iterations = 1 << 10;
//iterations = 50;
stringstream results;
vector<uint> currentPath, bestPath;
currentPath.reserve(vertexNumber), bestPath.reserve(vertexNumber);
//TabuArray tabuList(vertexNumber, tabuListSize);
TabuList tabuList(tabuListSize);
uint currentCost = 0, bestCost = 0, diverseLimit = (iterations >> 4) + 1, noChange = 0;
// 1 - Create an initial solution(could be created randomly), now call it the current solution.
for (uint i = 0; i < vertexNumber; ++i)
currentPath.push_back(i);
currentCost = calculateCost(currentPath);
bestPath = currentPath;
bestCost = currentCost;
for (uint i = 0; i < iterations; ++i)
{
// 2 - Find the best neighbor of the current solution by applying certain moves.
currentCost = getBestNeighbour(tabuList, currentPath);
// 3 - If the best neighbor is reached my performing a non - tabu move, accept as the new current solution.
// else, find another neighbor(best non - tabu neighbour).
if (currentCost < bestCost)
{
bestPath = currentPath;
bestCost = currentCost;
noChange = 0;
}
else
{
if (++noChange > diverseLimit)
{
random_shuffle(currentPath.begin(), currentPath.end());
currentCost = calculateCost(currentPath);
tabuList.clear();
noChange = 0;
}
}
}
// 4 - If maximum number of iterations are reached(or any other stopping condition), go to 5, else go to 2.
// 5 - Globally best solution is the best solution we found throughout the iterations.
double duration = (clock() - overallTime) / (double)CLOCKS_PER_SEC;
results << "Koszt drogi: " << bestCost << "\nCalkowity czas trwania: " << duration << " sekund\n";
return Data(bestPath, bestCost, results.str(), duration);
}
