# POJ 2442 Sequence （堆）

Sequence

 Time Limit: 6000MS Memory Limit: 65536K Total Submissions: 9316 Accepted: 3113

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

### AC 代码

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<iostream>
using namespace std;
#include<queue>
#include<map>

typedef __int64 LL;

int num[2001];
priority_queue<int,vector<int>,greater<int> >d; //小顶堆
priority_queue<int>q;       //大顶堆

int main()
{
int T;
scanf("%d",&T);
while(T--)
{
int m,n,x;
scanf("%d%d",&m,&n);
for(int i=0; i<n; i++)
{
scanf("%d",&x);
d.push(x);      //首先当只有一组的时候
}
for(int i=1; i<m; i++)  //添加剩下的组
{
for(int j=0; j<n; j++)
scanf("%d",num+j);
while(!d.empty())
{
x=d.top();
d.pop();
for(int j=0; j<n; j++)  //二重循环计算
{
if((int)q.size()<n)
q.push(x+num[j]);
else if(x+num[j]<q.top())
{
q.pop();
q.push(x+num[j]);
}
}
}
while(!q.empty())
{
d.push(q.top());
q.pop();
}
}
printf("%d",d.top());
d.pop();
while(!d.empty())
{
printf(" %d",d.top());
d.pop();
}
printf("\n");
}
return 0;
}