【POJ】 2442——Sequence【STL—优先队列】

题意；m * n 的矩阵 m行n列，每一行取一个数加起来，形成一个排列，升序输出排列和最小的前n个。

解题思路：维护两个数组，和一个优先队列，两个数组滚动使用。先对输入进来的每一行进行排序（升序）。把相邻的两行的值的和压入队列，在这个过程中队列中只保留n个元素，即比较小的n个， 多出n个就， 弹出最大的。

具体步骤如下：

1，读入第一行a[i]，把第一行进行排序，再读入第二行b[i]，用第一行的最小值a[0]与第二行的每个数相加，压入队列。

2，循环a[1]...a[n]与b[0]...b[n]的和与队列中的最大值进行比较，如果和值小于最大值就弹出最大值，压入和值，如果大于等于就break（因为是b[i]排过序的，这里和值如果大于等于，后面的和值也只能是大于等于）

3，按升序更新a[i]，a[i]就是队列中的值， 反复第2步，直到结束.

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<algorithm>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<list>
#include<iostream>
#include<map>
#include<queue>
#include<set>
#include<stack>
#include<vector>
#define MAX_N 2005
using namespace std;

int n, m;
int a[MAX_N];
int b[MAX_N];
priority_queue<int> pq;

int main()
{
    int t;
    scanf("%d", &t);
    while(t--)
    {
        scanf("%d%d", &m, &n);
        for(int i = 0 ; i < n; i++)
            scanf("%d", &a[i]);
        sort(a, a + n);
        while(--m)
        {
            for(int i = 0 ; i < n ; i++)
            {
                scanf("%d", &b[i]);
                pq.push(a[0] + b[i]);
            }
            sort(b, b + n);
            for(int i = 1 ; i < n ; i++)
            {
                for(int j = 0 ; j < n ; j++)
                {
                    if(a[i] + b[j] >= pq.top())
                        break;
                    pq.pop();
                    pq.push(a[i] + b[j]);
                }
            }
            for(int i = n -1 ; i >= 0 ; i--)
            {
                a[i] = pq.top();
                pq.pop();
            }
        }
        for(int i = 0 ; i < n - 1 ; i++)
        {
            printf("%d ", a[i]);
        }
        printf("%d\n", a[n -1]);
    }
    return 0;
}