Sequence
| Time Limit: 6000MS |
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Memory Limit: 65536K |
| Total Submissions: 7224 |
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Accepted: 2369 |
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
Source
POJ Monthly,Guang Lin
题解就是：两个数列各个数字分别相加，取前n个小的存下，成为一个新的数列，而后再取一个新的数列，重复上述步骤。
AC代码：
#include
#include
#include
using namespace std; priority_queue
qu; int a[2010]; int b[2010]; int main(){ int T; cin>>T; while(T--){ int n,m; cin>>m>>n; for(int i=1;i<=n;i++) cin>>a[i]; sort(a+1,a+n+1); m--; while(m--){ while(!qu.empty()) qu.pop(); for(int i=1;i<=n;i++){ cin>>b[i]; qu.push(b[i]+a[1]); } sort(b+1,b+n+1); for(int i=2;i<=n;i++) for(int j=1;j<=n;j++){ if(a[i]+b[j]
=1;i--){ a[i]=qu.top(); qu.pop(); } } for(int i=1;i
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