Sequence用堆排序

Description

Given m sequences, each contains n non-negative integer. Now we may select one number from each sequence to form a sequence with m integers. It's clear that we may get n ^ m this kind of sequences. Then we can calculate the sum of numbers in each sequence, and get n ^ m values. What we need is the smallest n sums. Could you help us?

Input

The first line is an integer T, which shows the number of test cases, and then T test cases follow. The first line of each case contains two integers m, n (0 < m <= 100, 0 < n <= 2000). The following m lines indicate the m sequence respectively. No integer in the sequence is greater than 10000.

Output

For each test case, print a line with the smallest n sums in increasing order, which is separated by a space.

Sample Input

1
2 3
1 2 3
2 2 3

Sample Output

3 3 4

 #include"iostream"
 #include"algorithm"
 #include"ctime"
 #include"cstdio"
 #include"cctype"
 using namespace std;
 #define maxx 2008
 int a[maxx],b[maxx],sum[maxx];
 int main()
 {
     int i,j,k,t,n,m,temp;
     scanf("%d",&t);
     while(t--)
     {
         scanf("%d%d",&m,&n);
         for(i=;i<n;i++)
           scanf("%d",&a[i]);
         for(i=;i<m;i++)
         {
             sort(a,a+n);
             for(j=;j<n;j++)
               scanf("%d",&b[j]);
             for(j=;j<n;j++)
               sum[j]=a[j]+b[];
             make_heap(sum,sum+n);
             for(j=;j<n;j++)
               for(k=;k<n;k++)
               {
                   temp=b[j]+a[k];
                   if(temp>=sum[])
                      break;
                   pop_heap(sum,sum+n);
                 sum[n-]=temp;
                 push_heap(sum,sum+n);
               }
             for(j=;j<n;j++)
                a[j]=sum[j];
         }
         sort(a,a+n);
         printf("%d",a[]);
         for(j=;j<n;j++)
            printf(" %d",a[j]);
         printf("\n");
         printf("time :%d\n",clock());
     }
     return ;
 }