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[],b[],i,j;
     priority_queue<int,vector<int>,less<int> >que;
     scanf("%d",&t);
     while(t--)
     {
         scanf("%d %d",&m,&n);
         m--;
         for(i=; i<n; i++)
             scanf("%d",&a[i]);
         sort(a,a+n);
         while(m--)
         {
             for(i=; i<n; i++)
             {
                 scanf("%d",&b[i]);
                 que.push(a[]+b[i]);//先进去N个
             }
             sort(b,b+n);//这个应该会用了
             for(i=; i<n; i++)
             {
                 for(j=; 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-; i>-; i--)
             {
                 a[i]=que.top();
                 que.pop();
             }
         }
         printf("%d",a[]);
         for(i=; i<n; i++)
             printf(" %d",a[i]);
         printf("\n");
     }
     return ;
 }

Hint

Huge input,scanf is recommended.