(Application of heap) POJ 2442

Sequence

Time Limit: 6000MS		Memory Limit: 65536K
Total Submissions: 7762		Accepted: 2565

Description

Given m sequences, each contains n non-negative integer. Now we may select one of the sequence to form a sequence with M integers. It's clear that we could get n ^ m this kind of sequences. Then we can calculate the sum of numbers in each sequence, and get n ^ m values. What are we need is the smallest n sums. Could help us?

Input

The first line is a integer T, which shows the number of test cases, and then T test cases follow. The first line of each case contains integers m, n (0 < M <=, 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

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", &AMP;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
　　
(Application of heap) POJ 2442