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<iostream>
#include<cstdio>
#include<queue>
#include<algorithm>
using namespace std;
int a[][],b[];
priority_queue<int>p;
int main()
{
int T;
scanf("%d",&T);
while(T--){
int n ,m;
scanf("%d%d",&m,&n); //m个序列，每个含n个数字
for(int i=;i<m;i++){
for(int j=;j<n;j++)scanf("%d",&a[i][j]);
sort(a[i],a[i]+n); //每个序列输入完后先排序
}
for(int i=;i<n;i++)
p.push(a[][i]);
for(int i=;i<m;i++){
for(int j=;j<n;j++){
b[j]=p.top();
p.pop();
}
for(int j=;j<n;j++){
for(int t=n-;t>=;t--){
if(j==)p.push(a[i][]+b[t]); //将第i个序列的第一项与加之i-1序列的前n小项相加
else{
if(p.top()>b[t]+a[i][j]){
p.pop();
p.push(a[i][j]+b[t]);
}
else break;
}
}
}
}
for(int i=;i<n;i++){
b[i]=p.top();
p.pop();
}
for(int i=n-;i>=;i--){
if(i==)printf("%d\n",b[i]);
else printf("%d ",b[i]);
}
}
//system("pause");
return ;
}