Sequence

最新推荐文章于 2024-03-30 12:15:47 发布

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最新推荐文章于 2024-03-30 12:15:47 发布

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分类专栏： STL

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本文介绍了一种算法，用于从给定的多个整数序列中选取元素形成新序列，目标是找到n个具有最小总和的新序列。算法首先对每个序列进行排序，然后使用堆数据结构来维护当前可能的最小和，通过迭代更新堆中的元素，最终得到所需的结果。

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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

Sample Output

3 3 4

1.将第一序列读入data1向量中，并按升序排序。

2.将数据读入data2向量中，并按升序排序。

将data2[0] + data1[i] ( 0<=i<=n-1)读入dataq向量中

用make_heap对dataq建堆。

然后data2[1] + data1[i] (0<=i<=n-1),如果data2[1] + data1[i]比堆dataq的顶点大,则退出，否则删除

堆的顶点，插入data2[1] + data1[i]。然后是data2[2],...data2[n - 1]

3.将dataq的数据拷贝到data1中，并对data1按升序排序

4.循环2,3步，直到所有数据读入完毕。

5.打印data1中的数据即为结果。

ac:

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<stdlib.h>
#include<queue>
#define MAX 2005
using namespace std;
int sum[MAX],data[MAX],heap[MAX];

int main()
{
    std::ios::sync_with_stdio(false);
    int i,j,k,t,m,n,temp;
    cin>>t;
    while(t--)
    {
        memset(sum,0,sizeof(sum));
        cin>>m>>n;
        for(int i=0;i<n;i++)
            cin>>sum[i];
        for(int i=1;i<m;i++)
        {
            memset(data,0,sizeof(data));
            sort(sum,sum+n);
            for(int j=0;j<n;j++)
                cin>>data[j];
            sort(data,data+n);
            for(int j=0;j<n;j++)
                heap[j]=sum[0]+data[j];
            make_heap(heap,heap+n);
            for(int j=1;j<n;j++)
            {
                for(int k=0;k<n;k++)
                {
                    temp=sum[j]+data[k];
                    if(temp>=heap[0])
                        break;
                    pop_heap(heap,heap+n);
                    heap[n-1]=temp;
                    push_heap(heap,heap+n);
                }
            }
            for(int j=0;j<n;j++)
                sum[j]=heap[j];
        }
        sort(sum,sum+n);
        for(int j=0;j<n-1;j++)
            cout<<sum[j]<<" ";
        cout<<sum[n-1]<<endl;
    }
    return 0;
}