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=0;i<n;i++)
          scanf("%d",&a[i]);
        for(i=1;i<m;i++)
        {
            sort(a,a+n);
            for(j=0;j<n;j++)
              scanf("%d",&b[j]);
            for(j=0;j<n;j++)
              sum[j]=a[j]+b[0];
            make_heap(sum,sum+n);
            for(j=1;j<n;j++)
              for(k=0;k<n;k++)
              {
                  temp=b[j]+a[k];
                  if(temp>=sum[0])
                     break;
                  pop_heap(sum,sum+n);
                sum[n-1]=temp;
                push_heap(sum,sum+n);   
              }   
            for(j=0;j<n;j++)
               a[j]=sum[j];   
        } 
        sort(a,a+n);
        printf("%d",a[0]);
        for(j=1;j<n;j++)
           printf(" %d",a[j]);
        printf("
");
        printf("time :%d
",clock());    
    }
    return 0;   
}