poj 2442 Sequence

题目链接：点击打开链接

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

题目大意：有M行　每行有n个数，每行拿出一个数来相加得和，共有n的m次方个和，求前N个最小的和
基本思路：用优先队列存储　每输入一行就对其进行加和判断存储，注意删数队列；
具体看代码

<span style="font-size:24px;">#include <iostream>
#include<queue>
#include<cstring>
#include<algorithm>
#include<cstdlib>
#include<cstdio>
using namespace std;
int main()
{
int t;
int n,m;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&m,&n);
int x;
priority_queue <int ,vector<int >,greater<int> >q;//从小到达存数
priority_queue <int ,vector<int >,less<int > >p;//从大到小存储
int a[2005];
for(int i = 0; i<n; i++)
{
scanf("%d",&x);//先将一行存入队列
q.push(x);
}
for(int i=1; i<m; i++)
{
for(int j=0; j<n; j++)
{
scanf("%d",&a[j]);
}
while(!q.empty())//将上一行的每个数拿出与当前行加和
{
int sum=q.top();
q.pop();
for(int j=0;j<n;j++)
{
if(p.size()>=n&&p.top()>sum+a[j])//因为只要前n个数　所以只保留前n个最小的就行
{
p.pop();
p.push(sum+a[j]);
}
else if(p.size()<n)
{
p.push(sum+a[j]);
}
}
}
while(!p.empty())
{
q.push(p.top());
p.pop();
}
}
for(int i=0;i<n;i++)
{
if(i==0)
{
printf("%d",q.top());
q.pop();
}
else
{
printf(" %d",q.top());
q.pop();
}
}
printf("\n");
}
return 0;
}
</span>