题意：给你n*m的矩阵，然后每行取一个元素，组成一个包含n个元素的序列，一共有n^m种序列，让你求出序列和最小的前n个序列的序列和;
我一开始的做法是用数组进行模拟这个堆的过程，但是由于每次排序函数的时间都很多，所以超时了啊、
后来看了一下题解用优先队列做的，这样判断一下如果大于队首元素就不如队列，可以省去很多的时间、、
Sequence
| Time Limit: 6000MS | Memory Limit: 65536K | |
| Total Submissions: 6246 | Accepted: 1945 |
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?
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.
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 <stdio.h>
#include <string.h>
#include <queue>
#include <iostream>
#include <stdlib.h>
using namespace std;
int a[40005];
int main()
{
int t, n, m, i, j, flat, sum;
scanf("%d",&t);
while(t--)
{
int x;
scanf("%d %d",&n, &m);
priority_queue<int , vector<int> , greater<int> >p;
priority_queue<int , vector<int> , less<int> >q;
for(i = 0; i < m; i++)
{
scanf("%d",&x);
p.push(x);
}
for(i = 1; i < n; i++)
{
for(j = 0; j < m; j++)
scanf("%d",&a[j]);
while(!p.empty())
{
sum = p.top();
p.pop();
for(j = 0; j < m; j++)
if(q.size() < m)
q.push(sum+a[j]);
else if(q.size() == m && q.top() > sum+a[j])
{
q.pop();
q.push(sum+a[j]);
}
}
while(!q.empty())
{
p.push(q.top());
q.pop();
}
}
flat = 0;
for(i = 0; i < m; i++)
{
if(flat)
printf(" ");
printf("%d",p.top());
p.pop();
flat = 1;
}
printf("\n");
}
return 0;
}