多机任务调度

多机任务调度

今天做http://202.120.80.191/problem.php?problemid=2588，开始把题目理解错了，原来每个进程的执行顺序是不可改变的。我没看到这一点。如果可以变，就成了多处理机多任务调度问题，看网上说这是NP问题。写了很久程序，才发现把题目理解错了。不过我觉得我的多处理机多任务调度的方法还是挺好的。
如果要把作业分成N份，先求出平均的处理时间（=总时间/N）。然后用动态规划求出小于等于该处理时间的一组作业。之后把剩余的作业堪称是分成N-1份，按照上面的方法做。直到求出N-1组小于等于平均处理时间的作业。则剩下的作业就是大于等于平均剩余时间的，求出其和就是结果。不知道效果怎么样，以后证明一下吧。

#include  < iostream >
#include  < algorithm >
#include  < vector >
#include  < string >
using   namespace  std;
const   int  maxn = 110 ;
int  m,n;
int  time[maxn];
void  dp( int  avg)
{
     int  i;
     int  pre[maxn][ 2 ],a[maxn][ 2 ];
     bool  visit[maxn];
     int  b[maxn];

    memset(visit, 0 , sizeof (visit));
    
    a[ 0 ][ 0 ] = a[ 0 ][ 1 ] = 0 ;
    pre[ 0 ][ 0 ] = pre[ 0 ][ 1 ] = 0 ;
    
     for (i = 1 ;i <= m;i ++ )
    {
         if (a[i - 1 ][ 1 ] !=- 1   &&  a[i - 1 ][ 1 ] > a[i - 1 ][ 0 ])
        {
            a[i][ 0 ] = a[i - 1 ][ 1 ];
            pre[i][ 0 ] = 1 ;
        }
         else
        {
            a[i][ 0 ] = a[i - 1 ][ 0 ];
            pre[i][ 0 ] = 0 ;
        }

         int  t1 = a[i - 1 ][ 0 ] + time[i] ,t2 = a[i - 1 ][ 1 ] + time[i];

        a[i][ 1 ] =- 1 ;
         if (t1 <= avg)
        {
            a[i][ 1 ] = t1;
            pre[i][ 1 ] = 0 ;
        }
         if (a[i - 1 ][ 1 ] !=- 1   &&  t2 <= avg  &&  t2 > t1)
        {
            a[i][ 1 ] = t2;
            pre[i][ 1 ] = 1 ;
        }
    }
    
     int  t;

     if (a[m][ 0 ] > a[m][ 1 ]) t = 0 ;
     else  t = 1 ;

     for (i = m;i > 0 ;i -- )
    {
         if (t == 1 )    visit[i] = true ;
        t = pre[i][t];
    }

     int  j;
     for (i = 1 ,j = 0 ;i <= m;i ++ )
         if ( ! visit[i])
            b[ ++ j] = time[i];
     for (i = 1 ;i <= j;i ++ )
        time[i] = b[i];
    m = j;
}
void  solve()
{
     int  i,j;
     int  count,sum,avg;
     for (i = 1 ;i < n;i ++ )
    {
        count = sum = 0 ;
         for (j = 1 ;j <= m;j ++ )
            sum += time[j];
        avg = sum / (n - i + 1 );
        dp(avg);
    }
     int  res = 0 ;
     for (i = 1 ;i <= m;i ++ )
        res += time[i];
    printf( " %d\n " ,res);
}
int  main()
{
     int  t;
    scanf( " %d " , & t);
     while (t -- )
    {
        scanf( " %d%d " , & m, & n);
         for ( int  i = 1 ;i <= m;i ++ )
            scanf( " %d " , & time[i]);
        solve();
    }
}

如果就本题而言，必须按顺序执行，只需二分即可。不是一般的简单

#include  < iostream >
#include  < algorithm >
#include  < vector >
#include  < string >
using   namespace  std;
const   int  maxn = 110 ;
int  time[maxn];
int  m,n;
bool  check( int  t)
{
     int  count = 1 ,total = 0 ;
     for ( int  i = 0 ;i < m;i ++ )
    {
         if (total + time[i] <= t)
            total += time[i];
         else
        {
            total = time[i];
            count ++ ;
        }
    }
     return  (count <= n);
}
int  main()
{
     int  t;
     int  low,up,mid;
    scanf( " %d " , & t);
     while (t -- )
    {
        scanf( " %d%d " , & m, & n);
        low = 0 ,up = 0 ;
         for ( int  i = 0 ;i < m;i ++ )
        {
            scanf( " %d " , & time[i]);
            up += time[i];
             if (time[i] > low) low = time[i];
        }
         while (low < up)
        {
            mid = (low + up) / 2 ;
             if (check(mid))
                up = mid;
             else  low = mid + 1 ;
        }
        printf( " %d\n " ,low);
    }
}