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Codeforces#590(1234)——B2Social Network (hard version)

运维开发网 https://www.qedev.com 2020-07-23 11:33 出处:网络 作者:运维开发网整理
B2. Social Network (hard version) time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output The only difference between easy and hard versions are c

B2. Social Network (hard version)

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

The only difference between easy and hard versions are constraints on n">nn and k">kk.

You are messaging in one of the popular social networks via your smartphone. Your smartphone can show at most k">kk most recent conversations with your friends. Initially, the screen is empty (i.e. the number of displayed conversations equals 0">00).

Each conversation is between you and some of your friends. There is at most one conversation with any of your friends. So each conversation is uniquely defined by your friend.

You (suddenly!) have the ability to see the future. You know that during the day you will receive n">nn messages, the i">ii-th message will be received from the friend with ID idi">idiidi (1≤idi≤109">1≤idi≤1091≤idi≤109).

If you receive a message from idi">idiidi in the conversation which is currently displayed on the smartphone then nothing happens: the conversations of the screen do not change and do not change their order, you read the message and continue waiting for new messages.

Otherwise (i.e. if there is no conversation with idi">idiidi on the screen):

  • Firstly, if the number of conversations displayed on the screen is k">kk, the last conversation (which has the position k">kk) is removed from the screen.
  • Now the number of conversations on the screen is guaranteed to be less than k">kk and the conversation with the friend idi">idiidi is not displayed on the screen.
  • The conversation with the friend idi">idiidi appears on the first (the topmost) position on the screen and all the other displayed conversations are shifted one position down.

Your task is to find the list of conversations (in the order they are displayed on the screen) after processing all n">nn messages.

Input

The first line of the input contains two integers n">nn and k">kk (1≤n,k≤2⋅105)">1≤n,k≤2⋅105)1≤n,k≤2⋅105) — the number of messages and the number of conversations your smartphone can show.

The second line of the input contains n">nn integers id1,id2,…,idn">id1,id2,…,idnid1,id2,…,idn (1≤idi≤109">1≤idi≤1091≤idi≤109), where idi">idiidi is the ID of the friend which sends you the i">ii-th message.

Output

In the first line of the output print one integer m">mm (1≤m≤min(n,k)">1≤m≤min(n,k)1≤m≤min(n,k)) — the number of conversations shown after receiving all n">nn messages.

In the second line print m">mm integers ids1,ids2,…,idsm">ids1,ids2,…,idsmids1,ids2,…,idsm, where idsi">idsiidsi should be equal to the ID of the friend corresponding to the conversation displayed on the position i">ii after receiving all n">nn messages.

Examples

input

 
7 2
1 2 3 2 1 3 2

output

 
2
2 1 

input

 
10 4
2 3 3 1 1 2 1 2 3 3

output

 
3
1 3 2 

Note

In the first example the list of conversations will change in the following way (in order from the first to last message):

  • []">[][];
  • [1]">[1][1];
  • [2,1]">[2,1][2,1];
  • [3,2]">[3,2][3,2];
  • [3,2]">[3,2][3,2];
  • [1,3]">[1,3][1,3];
  • [1,3]">[1,3][1,3];
  • [2,1]">[2,1][2,1].

In the second example the list of conversations will change in the following way:

  • []">[][];
  • [2]">[2][2];
  • [3,2]">[3,2][3,2];
  • [3,2]">[3,2][3,2];
  • [1,3,2]">[1,3,2][1,3,2];
  • and then the list will not change till the end. 

 这道题可以说我并不清楚我当时是否能写出来(因为我一开始并没有觉得写hard也算一题)不过后来直接看了题目的题解,就总结一下吧。

首先这道题是这样子的:如果某联系人已经在队列中,就啥也不做;如果不在,并且加入之前数量就等于k,最后一个联系人出队,再加进去新的联系人,如果小于k直接入队。

easy version真的简单啊,vector模拟:

#include <stdio.h>
#include <vector>
#include <set>
using namespace std;
vector<int> v;
int main(void)
{
	int n, k, t;
	scanf("%d %d", &n, &k);
	for(int i = 0; i < n; i++)
	{
		scanf("%d", &t); 
		vector<int>::iterator it = v.begin();
		for(it; it != v.end(); it++)//查找是否存在 
		{
			if(*it == t)
				break;
		}
		if(it == v.end())//不存在(存在就什么都不做) 
		{
			if(v.size() >= k)//size大于等于k 
			{
				v.erase(v.begin());//删掉最前面的(我反向模拟题目里面的规则,及我的是先进排在前面,当然出去也是从最前开始出) 
				v.push_back(t);//压入末尾 
			}
			else
				v.push_back(t);//size小于k 
		}
	}
	printf("%d\n", v.size());
	for(int i = v.size() - 1; i >= 0; i--)
	{
		printf("%d", v[i]);
		if(i)
			printf(" ");
	}
	return 0;
}
		

  当时对付hard version的2*105就肯定要超时了,所以采用STL一起上,queue模拟,set查找,vector最后存储,注意queue和set里面的元素一样,然后按顺序赋给vector。

#include <stdio.h>
#include <queue>
#include <set>
#include <vector>
#include <algorithm>
using namespace std;
queue<int> q;
set<int> s;
int main(void)
{
	int n, k, t;
	scanf("%d %d", &n, &k);
	for(int i = 0; i < n; i++)
	{
		scanf("%d", &t); 
		if(s.find(t) == s.end())//set没找到 
		{
			if(s.size() >= k)//数量超了 
			{
				int b = q.front();//记录要出去的元素 
				q.pop();//队列除去该元素 
				q.push(t);//队列进入新元素 
				s.insert(t);//set进入新元素 
				s.erase(b);//set删除要出去的元素 
			}
			else
			{
				q.push(t);
				s.insert(t);
			}
		}
	}
	printf("%d\n", s.size());
	vector<int> v;
	while(!q.empty()) 
	{
		v.push_back(q.front());
		q.pop();
	}	
	reverse(v.begin(), v.end());//因为queue的顺序和题目要求刚好相反,于是反向vector,因为queue不能反向 
	for(int i = 0; i < v.size(); i++)
	{
		printf("%d", v[i]);
		if(i != v.size() - 1)
			printf(" ");
	}
	return 0;
}
		
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