【K小堆问题】POJ 2442 Sequence

求最小序列和

最新推荐文章于 2021-07-05 16:03:29 发布

原创最新推荐文章于 2021-07-05 16:03:29 发布 · 225 阅读

0 ·

CC 4.0 BY-SA版权

K小堆专栏收录该内容

1 篇文章

订阅专栏

本文介绍了一个算法问题，即从给定的多个整数序列中选取元素形成新的序列，目标是找到这些新序列中和最小的n个。文章通过使用轮换数组和最小堆的数据结构，提出了一种高效的解决方案。

Sequence

Time Limit: 6000MS		Memory Limit: 65536K
Total Submissions: 12525		Accepted: 4103

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

Sample Output

3 3 4

Source

POJ Monthly,Guang Lin

在洛谷的合并序列中得到启发，对于两个数组取前k小和问题，用to数组来记录arr【i】与tmp数组的第几个数相加，并用最小堆维护。

两个数组的理解后，多个的就很好办了，把前两个数组的前K小值放到轮换数组中，与后续数组进行相同的操作即可

//#include <bits/stdc++.h>
#include <iostream>
#include <queue>
#include <cmath>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
//typedef __int128 bll;
const ll maxn = 2e3+100;
const ll mod = 1e9+7;
//const ld pi = acos(-1.0);
const ll inf = 1e18;
const ld eps = 1e-5;
//const ld e = exp(1);

ll T,n,m,arr[maxn],tmp[maxn],to[maxn]; //tmp轮换数组 

int main()
{	
    //ios::sync_with_stdio(false);
    //cin.tie(0),cout.tie(0);
	
	cin >> T;
	while(T--)
	{
		
		memset(arr,0,sizeof(arr));
		memset(tmp,0,sizeof(tmp));
		
		cin >> n >> m;
		
		for(ll i = 1; i <= n; i++)
		{
			memset(to,0,sizeof(to));
			
			priority_queue< pair<ll,ll> , vector < pair<ll,ll> >, greater< pair<ll,ll> > >Q;
			
			
			for(ll j = 1; j <= m; j++)
			{
				if(i == 1)
					cin >> tmp[j];
				else
				{
					cin >> arr[j];
				}
					
			}
			sort(tmp+1,tmp+1+m);
			sort(arr+1,arr+1+m);
			if(i >= 2)
			{
				for(ll j = 1; j <= m; j++)
				{
					to[j]=1;  //arr数组第j个和tmp数组哪个相加 
					Q.push( make_pair( tmp[1]+arr[j],j ) );
				}
				
				ll ok[maxn];  
				for(ll k = 1; k <= m; k++)
				{
					ok[k] = Q.top().first;
					ll t = Q.top().second;
					Q.pop();
					to[t]++;
					
					Q.push( make_pair( tmp[ to[t] ]+arr[t],t ) );
				}
				
					
				for(ll k = 1; k <= m; k++)
				{
					tmp[k] = ok[k];
				}
				
			}
		}
		
		for(ll i = 1; i <= m; i++)
				cout << tmp[i] << " ";
		cout << endl;
	}
	
	return 0;
}