Sequence

Sequence

Time Limit: 6000MS		Memory Limit: 65536K
Total Submissions: 7312		Accepted: 2415

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

给一个N行M列的矩阵，从每一行中拿出一个数进行相加，求其和中的前M小

<span style="font-size:18px;">#include <iostream>
#include<stdio.h>
#include<math.h>
#include<string.h>
#include<algorithm>
#include<queue>
#include<set>
#include<string>
using namespace std;
int main()
{
    priority_queue<int,vector<int>,less<int> >p;
    int n,m,i,j,a[2011],b[2011],t,k;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        for(i=0; i<m; i++)
            scanf("%d",&a[i]);
        sort(a,a+m);
        for(k=1; k<n; k++)
        {
            for(i=0; i<m; i++)
            {
                scanf("%d",&b[i]);
                p.push(a[0]+b[i]);
            }
            sort(b,b+m);
            for(i=1; i<m; i++)
            {
                for(j=0; j<m; j++)
                {
                    if(a[i]+b[j]>p.top())
                        break;
                    p.pop();
                    p.push(a[i]+b[j]);
                }
            }
            for(i=0; i<m; i++)
            {
                a[m-i-1]=p.top();
                p.pop();
            }
        }
        for(i=0; i<m-1; i++)
            printf("%d ",a[i]);
        printf("%d\n",a[m-1]);
    }
}

</span>