POJ 2442 Sequence

F -  Sequence

Time Limit:6000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit  Status  Practice  POJ 2442

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<stdio.h>
#include<string.h>
#include<iostream>
#include<queue>
#include<algorithm>

#define N 2101

using namespace std;



int main()
{
    int T,i,j,n,m;
    int a[N];
    scanf("%d",&T);
    while(T--)
    {
        int pp;
        priority_queue< int,vector<int>,greater<int> >q;
        priority_queue< int,vector<int>,less<int> >p;
        scanf("%d%d",&n,&m);
        for(i=0; i<m; i++)
        {
            scanf("%d",&pp);
            q.push(pp);
        }
        for(i=1; i<n; i++)
        {
            for(j=0; j<m; j++)
            {
                scanf("%d",&a[j]);
            }
            while(!q.empty())
            {
                int mm = q.top();
                q.pop();
                for(j=0; j<m; j++)
                {
                    if(p.size()<m)
                    {
                        p.push(mm+a[j]);
                    }
                    else if(p.size()==m && p.top()>mm+a[j])
                    {
                        p.pop();
                        p.push(mm+a[j]);
                    }
                }
            }
            while(!p.empty())
            {
                q.push(p.top());
                p.pop();
            }
        }
        for(i=0; i<m; i++)
        {
            if(i == 0)
            {
                printf("%d",q.top());
                q.pop();
            }
            else
            {
                printf(" %d",q.top());
                q.pop();
            }
        }
        printf("\n");
    }
    return 0;
}