补番计划

Time Limit : 4000/2000ms (Java/Other)   Memory Limit : 65535/32768K (Java/Other)

Total Submission(s) : 8   Accepted Submission(s) : 1

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Problem Description

Input

Output

Sample Input

2
7
0
1 [Sword_Art_Online]
1 [Your_Lie_in_April]
2 [sakura_sou_no_pet_na_kanojo]
0
0
0
6
1 [Your_Lie_in_April]
1 [Your_Lie_in_April]
2 [Is_the_order_a_rabbit]
3 [Is_the_order_a_rabbit]
0
0

Sample Output

-1
sakura_sou_no_pet_na_kanojo
Sword_Art_Online
Your_Lie_in_April
Your_Lie_in_April
-1

尽管没学过c++  。

。

感觉应该和java差点儿相同。。。

题意中由于要从队尾  队首增加 所以选用---------------------双端队列

为了推断是否看过某番 。为了番名的唯一性----------------set容器

剩下的就是模拟了

#include <iostream>
#include <string.h>
#include <stdio.h>
#include <queue>
#include <set>
using namespace std;
int main()
{
	set<string>s;
	deque<string>q;
	int ncase,n;
	scanf("%d",&ncase);
	while(ncase--)
	{
		s.clear();
		while(!q.empty())
		q.pop_back();
		cin>>n;
		for(int i=0;i<n;i++)
		{
			int x;
			cin>>x;
			if(x==0)
			{
				if(!q.empty())
				{
					string x1=q.front();q.pop_front();
					if(s.find(x1)==s.end())
					{
						cout<<x1<<endl;
						s.insert(x1);
					}
					else
					{
						cout<<-1<<endl;
					}
					
				}
				else
				{
					cout<<-1<<endl;
				}
			}
			if(x==1)
			{
				string x2;
				cin>>x2;
				q.push_back(x2.substr(1,x2.length()-2));
			}
			if(x==2)
			{
				string x2;
				cin>>x2;
				q.push_front(x2.substr(1,x2.length()-2));
			}
			if(x==3)
			{
				string x2;
				cin>>x2;
				x2=x2.substr(1,x2.length()-2);
				int count=q.size();
				while(count--)
				{
					string x3=q.front();q.pop_front();
					if(x3.compare(x2)!=0)
					{
						q.push_back(x3);
					}
				}
			}
		}
	
	}
	return 0;
}