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银行家算法C语言实现 第4页

更新时间:2008-7-14:  来源:毕业论文

银行家算法C语言实现 第4页

#define M 50
int allocation[M][M],need[M][M],available[M];
 int n,m,r;
main()
{void check();
 void print();
 int i,j,p=0,q=0;
 int req[M], allocation1[M][M],need1[M][M],available1[M];
 printf("Please input the sum of processes:");
 scanf("%d",&n);

 for(j=0;j<m;j++) 
  scanf("%d",&need[i][j]);

 printf("Available\n");
 for (i=0;i<m;i++)
 scanf("%d",&available[i]);
 print();
 check();
 if (r==1)
   {do {
        printf("\nplease input the NO. of process: ");
        scanf("%d",&i);
        printf("please input the resources of request,A B C:");
        for(j=0;j<m;j++)
        scanf("%d",&req[j]);
        p=0;
        q=0;
       for(j=0;j<m;j++)
       if(req[j]>need[i][j]) p=1;
       if(p)  printf("The resources of request have been beyond the max number needed!");
       else {
             for(j=0;j<m;j++)
             if(req[j]>available[j]) q=1;
             if(q)  printf("There are not enough available resources!");
             else {for(j=0;j<m;j++)
                    { available1[j]=available[j];
                      allocation1[i][j]=allocation[i][j];
                      need1[i][j]=need[i][j];
                      available[j]=available[j]-req[j];
                      allocation[i][j]=allocation[i][j]+req[j];
                      need[i][j]=need[i][j]-req[j];
                     }
                     print();
                     check();
                    if (r==0) 
                      {for (j=0;j<m;j++)
                      {available[j]=available1[j];
                       allocation[i][j]=allocation1[i][j];
                       need[i][j]=need1[i][j];
                       }
                     printf("return:\n");
                     print();
                      }
                  }
              }
     printf("\nDo you want to continue? y or n?");
     }while (getch()=='y');
   }
}
void check()
{int k,f,v=0,i,j;
 int work[M],a[M],finish[M];
 r=1;
 for(i=0;i<n;i++)
    finish[i]=0;
 for(i=0;i<m;i++)
    work[i]=available[i];
 k=n;
 do{
    for (i=0;i<n;i++)
       {if (finish[i]==0)
         {f=1;
          for (j=0;j<m;j++)
            if (need[i][j]>work[j])
             f=0;
               if (f==1)
               {finish[i]=1;
                a[v++]=i;
                for (j=0;j<m;j++)
                    work[j]=work[j]+allocation[i][j];
               }
      }
    }
  k--;
   }while(k>0);
 f=1;
 for (i=0;i<n;i++)
    {
  if (finish[i]==0)
    {
     f=0;
     break;
    }
 }
 if (f==0)
 {
  printf("This is unsafe  \n");
     r=0;
    }
 else
 {
  printf("This is safe and the safe number is:");
  for (i=0;i<n;i++)
  printf("%d ",a[i]);
    }
}
void print()
 { int i,j;
   int process[M];
  printf("Process\t Allocation\t  Need\n");
  for (i=0;i<n;i++)
      process[i]=i;
        printf("\n");
    }
  printf("Available\n");
  for(i=0;i<m;i++)
  printf("%2d ",available[i]);
  printf("\n");
 }

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