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CMatchelo

Untitled

Apr 9th, 2017
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  1. int proximoPros(vector<Process>& procVec, int tam, int tempo)
  2. {
  3. int prox=0, maxAtual=9999;
  4. std::cout << tempo << std::endl;
  5. for (int i=0; i<tam; i++)
  6. {
  7. if (procVec[i].ent <= tempo && procVec[i].done==false)
  8. {
  9. if (procVec[i].dur < maxAtual)
  10. {
  11. prox=i;
  12. maxAtual = procVec[i].dur;
  13. std::cout << "proximo atual sera " << prox << " de duracao " << procVec[i].dur << std::endl;
  14. std::cout << "tempo de execucao = " << tempo << std::endl;
  15. }
  16. }
  17. }
  18. return prox;
  19. }
  20.  
  21. int sjf (vector<Process>& procVec, int tam)
  22. {
  23. double rettot = 0, esptot = 0, reptot = 0;
  24. int tempo=0,espera=0,retorno=0;
  25. int prox = -1;
  26. for (int i=0; i<tam; i++)
  27. {
  28. prox = proximoPros(procVec, tam, tempo);
  29. if (prox >= 0)
  30. {
  31. std::cout << prox << " executado" << std::endl;
  32. procVec[prox].done = true;
  33. tempo = tempo + procVec[prox].dur;
  34. prox = -1;
  35. }
  36. else
  37. {
  38. std::cout << "nenhum entrou" << std::endl;
  39. tam++;
  40. tempo+=tempo;
  41. }
  42. }
  43. std::cout << tempo << std::endl;
  44. }
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