Sequence(優先キュー、またはスタック)

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

全体的なアルゴリズム:
2行ごとのn個の最小値の和を求め,最後の行まで求める.

#include<iostream>
#include<algorithm>
#include<vector>

using namespace std;

int main()
{
	int times;
	cin >> times;
	while (times--)
	{
		int m, n;
		cin >> m >> n;
		vector<int>res(n);
		for (int i = 0; i < n; i++)
			cin >> res[i];
		
		vector<int>newl(n);
		
		int cop = m-1;
		while (cop--)
		{		
			for (int i = 0; i < n; i++)
				cin >> newl[i];
			sort(newl.begin(), newl.end());
			make_heap(res.begin(), res.end());       //make_heap       
			vector<int>oldl(res.begin(), res.end());
			sort(oldl.begin(), oldl.end());
			for (int i = 0; i < n; i++)
				res[i] += newl[0];     //        
			
			for (int w = 1; w < n; w++)
			{
				bool ex = true;
				for (int q =0; q<n ; q++)
				{
					int tem = newl[w] +oldl[q];        //   ，          
					if (tem < res[0])
					{
						ex = false;
						pop_heap(res.begin(), res.end()); //        
						res.pop_back();                   //    
						res.push_back(tem);
						push_heap(res.begin(), res.end()); //           
					}
					else
						break;
				}
				if (ex)
					break;
			}
			
		}
		sort(res.begin(), res.end());
		for (int i = 0; i != n-1; i++)
		{
			cout <<res[i] << ' ';
		}
		cout <<res[n - 1] << endl;
	}
	
}

もちろん優先キューでやることもできます.