STL--F-sequence (N * m-> calculate the smallest first m and)

F-Sequence Time limit:6000 ms Memory limit:65536kb 64bit Io format:% I64d & % i64usubmit status

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. cocould 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

12 31 2 32 2 3

Sample output

3 3 4

Thanks http://www.cnblogs.com/372465774y/archive/2012/07/09/2583866.html

First, let's think about two questions.

From these two results, we need to obtain m values for each array of N arrays, and find the smallest m values for the sum of array 1 and array 2, and then sum with array 3 to find the smallest m values, finally, we get the smallest m of all arrays.

Since each array is ordered, and we require the smallest M, array a [I] [J] + values in the queue> the first value of the queue, then, the value after a [I] [J] And the queue will be greater than the first, which is meaningless for the minimum m values to be solved. The queue stores the smallest m Sum of the current array to all previous arrays, and constantly updates the queue.

#include <cstdio>#include <cstring>#include <queue>#include <algorithm>using namespace std;int a[110][2100] , b[2100] ;priority_queue <int> p ;int main(){    int i , j , k , n , m , t ;    scanf("%d", &t);    while(t--)    {        scanf("%d %d", &n, &m);        for(i = 0 ; i < n ; i++)        {            for(j = 0 ; j < m ; j++)                scanf("%d", &a[i][j]);            sort(a[i],a[i]+m);        }        for(i = 0 ; i < m ; i++)            p.push(a[0][i]) ;        for(i = 1 ; i < n ; i++)        {            for(j = 0 ; j < m ; j++)            {                b[j] = p.top();                p.pop();            }            for(j = 0 ; j < m ; j++)            {                for(k = m-1 ; k >= 0 ; k--)                {                    if(j == 0)                        p.push( a[i][j]+b[k] );                    else                    {                        if( a[i][j] + b[k] < p.top() )                        {                            p.pop();                            p.push(a[i][j]+b[k]);                        }                        else                            break;                    }                }            }        }        for(j = 0 ; j < m ; j++)        {            b[j] = p.top();            p.pop();        }        for(j = m-1 ; j >= 0 ; j--)        {            if(j == 0)                printf("%d\n", b[j]);            else                printf("%d ", b[j]);        }    }    return 0;}