U - stl 的 优先队列 Ⅰ

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<iostream>
#include<cstdio>
#include<queue>
#include<algorithm>
using namespace std;
int a[][],b[];
priority_queue<int>p;
int main()
{
    int T;
    scanf("%d",&T);
    while(T--){
        int n ,m;
        scanf("%d%d",&m,&n);           //m个序列，每个含n个数字
        for(int i=;i<m;i++){
            for(int j=;j<n;j++)scanf("%d",&a[i][j]);
            sort(a[i],a[i]+n);                         //每个序列输入完后先排序
        }
        for(int i=;i<n;i++)
            p.push(a[][i]);
        for(int i=;i<m;i++){
            for(int j=;j<n;j++){
                b[j]=p.top();
                p.pop();
            }
            for(int j=;j<n;j++){
                for(int t=n-;t>=;t--){
                    if(j==)p.push(a[i][]+b[t]);    //将第i个序列的第一项与加之i-1序列的前n小项相加
                    else{
                        if(p.top()>b[t]+a[i][j]){
                        p.pop();
                        p.push(a[i][j]+b[t]);
                        }
                        else break;
                    }
                }
            }
        }
        for(int i=;i<n;i++){
            b[i]=p.top();
            p.pop();
        }
        for(int i=n-;i>=;i--){
            if(i==)printf("%d\n",b[i]);
            else printf("%d ",b[i]);
        }
    }
    //system("pause");
    return ;
}