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Sequence用堆排序

2024-07-06 04:06:21   220人阅读

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
 1 #include"iostream"
 2 #include"algorithm"
 3 #include"ctime"
 4 #include"cstdio"
 5 #include"cctype"
 6 using namespace std;
 7 #define maxx 2008
 8 int a[maxx],b[maxx],sum[maxx];
 9 int main()
10 {
11     int i,j,k,t,n,m,temp;
12     scanf("%d",&t);
13     while(t--)
14     {
15         scanf("%d%d",&m,&n);
16         for(i=0;i<n;i++)
17           scanf("%d",&a[i]);
18         for(i=1;i<m;i++)
19         {
20             sort(a,a+n);
21             for(j=0;j<n;j++)
22               scanf("%d",&b[j]);
23             for(j=0;j<n;j++)
24               sum[j]=a[j]+b[0];
25             make_heap(sum,sum+n);
26             for(j=1;j<n;j++)
27               for(k=0;k<n;k++)
28               {
29                   temp=b[j]+a[k];
30                   if(temp>=sum[0])
31                      break;
32                   pop_heap(sum,sum+n);
33                 sum[n-1]=temp;
34                 push_heap(sum,sum+n);   
35               }   
36             for(j=0;j<n;j++)
37                a[j]=sum[j];   
38         } 
39         sort(a,a+n);
40         printf("%d",a[0]);
41         for(j=1;j<n;j++)
42            printf(" %d",a[j]);
43         printf("\n");
44         printf("time :%d\n",clock());    
45     }
46     return 0;   
47 }
View Code

 


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