poj2442 Sequence

Sequence

Time Limit: 6000MS		Memory Limit: 65536K
Total Submissions: 8887		Accepted: 2953

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

题目大意：给你m个序列，每次从每个序列里挑选一个数组成一个新的序列，每个序列进行加和组成一个新的序列，然后输出前n小的序列。

对于两个序列我们就可以a[0]+b[0],a[1]+b[0]一直到a[n-1]+b[0],然后是依次加a[1]，这样我们就可以得到一个新的序列，如果只有两个序列，那我们的序列就是符合条件的，如果对于三个序列那就是两个序列组成的新序列看成新的a序列，第三个序列就是b序列，想一想，是不是？

这样依次迭代，到最后就是所求的m个序列组成的新序列，如果我们全部暴力存储，显然时间和空间开销都是非常大的，到最后跟直接m个序列暴力是没有区别的，达到n^m。

╮(╯▽╰)╭人家也没让你求辣么多啊，每次维护长为n的数组就可以啦

这里我用一个优先队列来维护哒464ms才

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include