(堆的应用) poj 2442

Sequence

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

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<string>
#include<algorithm>
#include<cstdlib>
#include<queue>
#include<vector>
#include<stack>
using namespace std;
int tt,m,n;
int a[2010],b[2010],heap[2010];
int main()
{
    scanf("%d",&tt);
    while(tt--)
    {
        scanf("%d%d",&m,&n);
        for(int i=0;i<n;i++)
            scanf("%d",&a[i]);
        sort(a,a+n);
        m--;
        while(m--)
        {
            for(int i=0;i<n;i++)
                scanf("%d",&b[i]);
            sort(b,b+n);
            for(int i=0;i<n;i++)
                heap[i]=a[i]+b[0];
            make_heap(heap,heap+n);
            for(int i=1;i<n;i++)
            {
                bool ok=0;
                for(int j=0;j<n;j++)
                {
                    int temp=a[j]+b[i];
                    if(temp<heap[0])
                    {
                        pop_heap(heap,heap+n);
                        heap[n-1]=temp;
                        push_heap(heap,heap+n);
                        ok=1;
                    }
                    else
                        break;
                }
                if(!ok)
                    break;
            }
            for(int i=0;i<n;i++)
                a[i]=heap[i];
            sort(a,a+n);
        }
        for(int i=0;i<n-1;i++)
            printf("%d ",a[i]);
        printf("%d\n",a[n-1]);
    }
    return 0;
}