poj 2442 - c++编程基础 - 编程开发网

Sequence

Time Limit: 6000MS		Memory Limit: 65536K
Total Submissions: 7224		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

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