Sequence

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 
  
题目大意：输入m个集合，每个含n个数，求从每个集合取一个数后，按非降序输出前n小的和。
  
这个题是看了网上的题解才知道怎么做的
  
1：向A读入一个集合
  
2：向B读入下一个集合，用它最小的数遇上一个集合的每一个数相加，加入大根堆中，取出它的第二小的值与第一个集合中的值分别相加，如果该值小于大根堆堆顶的值，删除堆顶，将当前值加入，然后依此类推。
  
3：将堆中元素依次取出，逆向放入A中
  
4：重复2,3
  
//表达能力比较差……说得乱七八糟……
  
  
  
   
#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
#include<algorithm>
using namespace std;
priority_queue <long> q;
long a[2020], b[2020];
long t, n, m;


void init()
{

}
int main()
{
    freopen("poj2442.in", "r", stdin);
    scanf("%d",&t);
    for (long i = 1; i <= t; i++)
    {
        memset(a, sizeof(a), 0);
        scanf("%d%d", &n, &m);
        for (long i = 0; i < m; i++)
        {
            scanf("%d", &a[i]);
        }
        sort(a, a + m);
        for (long i = 2; i <= n; i++)
        {
            for (long j = 0; j <m; j++)
                scanf("%d", &b[j]);
            sort(b, b + m);
            for (long j = 0; j < m; j++)
                q.push(a[0]+b[j]);
            for (long j = 1; j < m; j++ )
                for (long k = 0; k < m; k++)
                {
                    if (a[j] + b[k] < q.top())
                    {
                        q.pop();
                        q.push(a[j] + b[k]);
                    }
                }

            for (long j = m - 1; j >= 0; j--)
            {
                a[j] = q.top();
                q.pop();
            }
        }

        for (long j = 0 ; j < m; j++ )
            printf ("%d ",a[j]);
        cout<<endl;
    }
    return 0;
}