POJ 1018 Communication System

//此题是DP问题，可是我一直没有想出公式
//组后只能根据带宽来枚举，再采用贪心策略

 

#include < iostream >
using   namespace  std;

int  main()
{
     int  a[ 101 ][ 202 ],num[ 101 ],n,i,j,k,h,temp,tt;
     int  minb,minp,sump;
     float  bp,max;
    cin >> tt;
     while (tt -- )
    {
        max = 0 ;
        scanf( " %d " , & n);
         for (i = 0 ;i < n; ++ i)
        {
            scanf( " %d " , & num[i]);
             for (j = 0 ;j < num[i]; ++ j)
                scanf( " %d%d " , & a[i][j], & a[i][j + 100 ]);
        }
         for (i = 0 ;i < n; ++ i)
             for (j = 0 ;j < num[i]; ++ j)
            {
                minb = a[i][j];
                sump = a[i][j + 100 ];
                 for (k = 0 ;k < n; ++ k)
                {
                     if (k == i)  continue ;
                    minp = 100000 ;
                     for (h = 0 ;h < num[k]; ++ h)
                         if (a[k][h] >= minb && minp > a[k][h + 100 ])
                            minp = a[k][h + 100 ];
                    sump += minp;
                     if (minp == 100000 )
                         break ;
                }
                 if (minp == 100000 )  break ;
                bp = float (minb) / sump;
                 if (bp > max) max = bp;
            }
            printf( " %.3f\n " ,max);
    }
    
     return   1 ;
}