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

12 31 2 32 2 3

Sample Output

3 3 4

#include<stdio.h>#include<string.h>#include<iostream>#include<queue>#include<algorithm>#define N 2101using 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;}