Hdu3527 (Hangdian) spy problem

Need to know ：vector See the following blog for basic details

STL application （ unfinished _m0_62044244 The blog of -CSDN Blog

Title Description

The NationalIntelligence( Intelligence work ) Council( The committee ) of X Nation receives a piece ofcredible( reliable ) informationthat Nation Y will send spies( The spy ) to stealNation X’s confidential( Confidential ) paper. So thecommander( The commander ) of TheNational Intelligence Council take measures immediately, he will investigate( survey ) people who will come into Nation X. Atthe same time, there are two List in the Commander’s hand, one is full of spiesthat Nation Y will send to Nation X, and the other one is full of spies thatNation X has sent to Nation Y before. There may be some overlaps( repeat ) of the two list. Because the spy( The spy ) may act two roles at the same time, whichmeans that he may be the one that is sent from Nation X to Nation Y, we justcall this type a “dual-spy”. So Nation Y may send “dual_spy” back to Nation X,and it is obvious now that it is good for Nation X, because “dual_spy” maybring back NationY’s confidential paper without worrying to be detention( Detainment ) by NationY’s frontier( The border ) So the commander decides to seize thosethat are sent by NationY, and let the ordinary people and the “dual_spy” in atthe same time .So can you decide a list that should be caught by the Commander? 
A:the list contains that will come to the NationX’s frontier. 
B:the list contains spies that will be sent by Nation Y. 
C:the list contains spies that were sent to NationY before. 

input

 There areseveral test cases. 
Each test case contains four parts, the first part contains 3 positive integersA, B, C, and A is the number which will come into the frontier. B is the numberthat will be sent by Nation Y, and C is the number that NationX has sent toNationY before. 
The second part contains A strings, the name list of that will come into thefrontier. 
The second part contains B strings, the name list of that are sent by NationY. 
The second part contains C strings, the name list of the “dual_spy”. 
There will be a blank line after each test case. 
There won’t be any repetitive( Repetitive ) names in asingle list, if repetitive names appear in two lists, they mean the samepeople. 
 

output

 Output the listthat the commander should caught (in the appearance order of the lists B).if noone should be caught, then , you should output “No enemy spy”
 

sample input

8 4 3

Zhao Qian Sun Li Zhou Wu Zheng Wang

Zhao Qian Sun Li

Zhao Zhou Zheng

2 2 2

Zhao Qian

Zhao Qian

Zhao Qian

sample output

Qian Sun Li

No enemy spy

ans 

#include<bits/stdc++.h>
using namespace std;
const int N=1e7+10;
vector<string> x,y,z,ans;// Deposit in the three line list   And a list of arrests 
int main()
{
	int a,b,c,i;string s;
	while(cin>>a>>b>>c)
	{
		x.clear(),y.clear(),z.clear(),ans.clear();
		for(i=0;i<a;i++)
		{
			cin>>s;x.push_back(s);
		}
		for(i=0;i<b;i++)
		{
			cin>>s;y.push_back(s);
		}
		for(i=0;i<c;i++)
		{
			cin>>s;z.push_back(s);
		}// The above step is to deposit in the three line list 
		for(i=0;i<b;i++)
		{// From the second line, find out what the first line has but the third line does not have, that is, the spy to be captured 
			if(find(x.begin(),x.end(),y[i])!=x.end())
			if(find(z.begin(),z.end(),y[i])==z.end())
			{
				ans.push_back(y[i]);
			}
		}
			if(!ans.size())
		{
			cout<<"NO enemy spy";
		}else{
			for(i=0;i<ans.size();i++)
			{
				cout<<ans[i]<<" ";
			}
		}
		cout<<endl;
	}
	return 0;
}