Sequence
| Time Limit: 6000MS |
|
Memory Limit: 65536K |
| Total Submissions: 7447 |
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Accepted: 2451 |
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
题意：给m个长度为n的数组，每次从每个数组中取一个数组成一个长度为m的数组（将数组中的数相加求和），操作n次，即取前n小和。
首先输入第一行（即第一个数组），升序排序，然后接下来输入m-1行，每次输入一行，将第一次输入的数组整体加上这次输入的第一个元素，然后维护成一个最大堆，再然后就要接着筛选，本质是继续维护这个最大堆，一直处理到输入结束即可。
#include
#include
#include
#include
#include
using namespace std; const int maxn=2010; int tem[maxn],ans[maxn],a[maxn]; int main() { int t,n,m; scanf("%d",&t); while(t--) { scanf("%d%d",&m,&n); for(int i=0;i
=tem[0])break; pop_heap(tem,tem+n); tem[n-1]=x; push_heap(tem,tem+n); } sort_heap(tem,tem+n); for(int i=0;i