Sequence《优先队列》

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

这个题的意思不是很难理解，关键是思想，一点点算肯定超时了；

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<queue>
using namespace std;
int main()
{
    int m,n,t,a[2005],b[2005],i,j;
    priority_queue<int,vector<int>,less<int> >que;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d %d",&m,&n);
        m--;
        for(i=0; i<n; i++)
            scanf("%d",&a[i]);
        sort(a,a+n);
        while(m--)
        {
            for(i=0; i<n; i++)
            {
                scanf("%d",&b[i]);
                que.push(a[0]+b[i]);//先进去N个
            }
            sort(b,b+n);//这个应该会用了
            for(i=1; i<n; i++)
            {
                for(j=0; j<n; j++)
                {
                    if(a[i]+b[j]>que.top())//因为有sort排序，所以，后面只会更大，可以break；
                        break;
                    que.pop();
                    que.push(a[i]+b[j]);
                }
            }
            for(i=n-1; i>-1; i--)
            {
                a[i]=que.top();
                que.pop();
            }
        }
        printf("%d",a[0]);
        for(i=1; i<n; i++)
            printf(" %d",a[i]);
        printf("
");
    }
    return 0;
}

Hint

Huge input,scanf is recommended.