poj 2442 -- Sequence

Sequence

Time Limit: 6000MS		Memory Limit: 65536K
Total Submissions: 7018		Accepted: 2265

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

思路：
　　　　1.s1[M] 表示第1组n个数，按递增排序。s2[M]表示第2组n个数，按递增排序
　　　　2.sum[i] = s2[0] + s1[i]  i ∈ 0，1,2,3……n-1  并将sum[]数组建最大堆。这里我用优先队列实现
　　　　3.计算ans = s2[i] + s1[j] i ∈ 1,2,3……n-1,   j ∈ 0，1,2,3……n-1  . 其中当i = 1时，j依次取j的范围，i = 2时，j依次取j的范围  …… 
　　　　4.每次计算的 ans >= 堆顶 则退出此次，进行下一次，否则的话将堆顶删除将ans加到堆中。
　　　　5.将堆转换成s1数组，继续输入第三组到s2数组。重复上述步骤。 则最后堆里的内容就是最后要求的结果。

/*======================================================================
 *           Author :   kevin
 *         Filename :   Sequence.cpp
 *       Creat time :   2014-07-23 09:27
 *      Description :
========================================================================*/
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <queue>
#include <cmath>
#define clr(a,b) memset(a,b,sizeof(a))
#define M 2005
using namespace std;
int s1[M],s2[M],sum[M];
priority_queue<int>que;
int main(int argc,char *argv[])
{
    int cas;
    scanf("%d",&cas);
    while(cas--){
        int n,m;
        scanf("%d%d",&n,&m);
        for(int i = 0; i < m; i++){
            scanf("%d",&s1[i]);
        }
        sort(s1,s1+m);
        for(int i = 1; i < n; i++){
            for(int j = 0; j < m; j++){
                scanf("%d",&s2[j]);
            }
            sort(s2,s2+m);
            for(int j = 0; j < m; j++){
                sum[j] = s2[0] + s1[j];
                que.push(sum[j]);
            }
            int mmax = 0;
            for(int k = 1; k < m; k++){
                for(int j = 0; j < m; j++){
                    mmax = s2[k] + s1[j];
                    if(mmax >= que.top()) break;
                    que.pop();
                    que.push(mmax);
                }
            }
            int cnt = 0;
            while(!que.empty()){
                s1[cnt++] = que.top();
                que.pop();
            }
            sort(s1,s1+cnt);
        }
        for(int i = 0; i < m; i++){
            if(i)
                printf(" %d",s1[i]);
            else printf("%d",s1[i]);
        }
        printf("
");
    }
    return 0;
}