#include    “stdio.h”

int     Finish [10] = { 0 };                                                        /*  进程最多10个  */
int     Max [10][10] = { 0 };                                                       /*  资源数最多10个  */
int     Alloc [10][10] = { 0 };
int     Need [10][10] = { 0 };
int     Available [10] = { 10 };
int     Request [10][10] = { 0 }; purse hanger
int       main ()
{
         
          int  n,m;
          int  i,j;
          int  c=1;
         
          printf(“\n输入进程个数和资源种类数 [ 都不超过10 ]\n “);
          scanf(“%d%d”, &n, &m );
         
         
         
          printf(“\n输入每类资源的数量\n”);
          for( i=0; i<= m-1; i++ )
          {
               scanf(“%d”, & Available [i]);
          }   
         
          printf(“\n输入每个进程对每类资源的最大需求量\n”);
          for( i=0; i<= n-1; i++ )
          {
               for( j=0; j<= m-1; )
               {
                    scanf(“%d”, & Max [i][j]);
                    if( Max [i][j] <= Available [j] )    {   Need [i][j] = Max [i][j];      j++;     }
                    else    { printf(“错误!重输\n”); }
               }
          }
         
          Need[10][10]=Max[10][10];
         
          Print( n, m );
          while( c )
          {
                
                 Allocation ( n, m );
                 Print( n, m );
                 printf(“\n如果打印区空白务必按0退出  否则按1继续分配资源\n”);        /* 打印区空白按1会导致死机 */
                 scanf(“%d”,&c);
          }
          return 0;
}

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/*  资源预分配模块  */

int    Allocation ( int n, int  m )
{
      
       int    i,j;
      
       printf(“\n输入每个进程被分配的每类资源量 \n “);
      
       for( i=0; i<= n-1; i++ )
       {
               for( j=0; !Finish [i] && j<= m-1; )
               {
                    scanf(“%d”, & Request [i][j]);
                    if( Request [i][j] <= Need [i][j] )    j++; 
                    else    { printf(“错误!重输\n”); }
               }
       }
       for( i=0; i<= n-1; i++ )
       {
            for( j=0; !Finish [i] && j<= m-1; j++ )
            {
                  Available [j] -= Request [i][j];
                  Need [i][j] -= Request [i][j];
                 
            }
       }
       Deal (  n, m );
     
       return  0;
}

/*  校验预分配模块  */

int    Deal(  int  n, int  m )
{
     
        int   Tag = 0;
        int   i,j,k;
       
        for( i=0; i<= n-1; i++ )
        {

            for( j=0; !Finish [i] && j<= m-1; j++ )
            {
                
                 if( Need [i][j]> Available [j] )          break;
                
            }
           
            if( j == m )                                                                 /* 有进程可以撤退  */
            {
                for( k=0; k<= n-1; k++ )                                 
                {
                     for( j=0; !Finish [k] && j<= m-1; j++ )
                     {
                          Alloc [k][j] += Request [k][j];
                     }
                }
                Finish [i] = 1;                                                          /* i记录撤退进程  */
                Tag = 1;                                             
                for(j=0; j<= m-1; j++)    Available [j] += Alloc [i][j];
               
                break;
                               
            }
       }
           
            if( !Tag )                                                               /*  无进程可以撤退  还原相应矩阵 */
            {
                for( i=0; i<= n-1; i++ )
                {
                     for( j=0; !Finish [i] && j<= m-1; j++ )
                     {
                          Available [j] += Request [i][j];
                          Need [i][j] += Request [i][j];
                 
                     }
                }
                Allocation ( n, m );
            }

      
       return  0;
}

/*   打印进程对资源利用情况   */

int    Print( int n, int  m )
{
       int    i,j;

       printf(“\n进程对资源利用情况如下：\n”);
       for( i=0; i<= n-1; i++ )
       {
           
            for( j=0; !Finish [i] && j<= m-1; j++ )
            {
                 printf(“\n%d <- %d[ %d ]  %d -> %d[ %d ]\t”,i, j, Alloc [i][j], i, j, Need [i][j]);
            }
           
       }
      
       return  0;
}