hdu3527 SPY 水题

SPY

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 531 Accepted Submission(s): 245

Problem Description
The National Intelligence Council of X Nation receives a piece of credible information that Nation Y will send spies to steal Nation X’s confidential paper. So the commander of The National Intelligence Council take measures immediately, he will investigate people who will come into NationX. At the same time, there are two List in the Commander’s hand, one is full of spies that Nation Y will send to Nation X, and the other one is full of spies that Nation X has sent to Nation Y before. There may be some overlaps of the two list. Because the spy may act two roles at the same time, which means that he may be the one that is sent from Nation X to Nation Y, we just call 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” may bring back NationY’s confidential paper without worrying to be detention by NationY’s frontier So the commander decides to seize those that are sent by NationY, and let the ordinary people and the “dual_spy” in at the 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 are several test cases.
Each test case contains four parts, the first part contains 3 positive integers A, B, C, and A is the number which will come into the frontier. B is the number that will be sent by Nation Y, and C is the number that NationX has sent to NationY before.
The second part contains A strings, the name list of that will come into the frontier.
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 names in a single list, if repetitive names appear in two lists, they mean the same people.

Output
Output the list that the commander should caught (in the appearance order of the lists B).if no one 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

题目翻译:
X国的国家情报局得到可靠的消息说Y国派来间谍要偷取X国的机密文件。局长立刻采取措施，对最近进入X国的人群进行调查。另外，局长手里还有两份名单，一份是Y国派来X国的间谍名单，另一份是以前X国派到Y国的间谍名单，这两份名单可能有重叠，因为有双重间谍存在，所以Y国可能会把X国派去的间谍再次派回来，显然这对X国是有好处的，他们可以明目张胆的带着Y国的机密文件回来，而不被边境扣押。现在局长决定在边境就抓获Y国派来的间谍，把平民和X国派过去的双重间谍放进过境，那
么他应该抓捕那些人呢？
A：入境人员名单
B：Y国派来的间谍名单
C：X国派过去的卧底名单.
没啥好说的, 看代码就好了.

#include<cstdio>
#include<cstring>
#include<string>
#include<sstream>
#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
vector<string> a,b,c;
char buf[999];
int main(){
    //freopen("hdu3527.in", "r", stdin);
    //freopen("hdu.out", "w", stdout);
    int i,j,A,B,C;
    while(cin>>A>>B>>C){
        a.clear();b.clear();c.clear();
        int flag=1;
        while(A--){
            cin>>buf;
            a.push_back(buf);
        }
        while(B--){
            cin>>buf;
            b.push_back(buf);
        }
        while(C--){
            cin>>buf;
            c.push_back(buf);
        }
        vector<string>::iterator it;
        for(it=b.begin();it!=b.end();it++){
            if(find(c.begin(),c.end(),*it)!=c.end())
                continue;
            if(find(a.begin(),a.end(),*it)!=a.end()){
                if(1==flag)
                    flag=0;
                else
                    cout<<" ";
                cout<<*it;
            }
        }
        if(1==flag)
            cout<<"No enemy spy";
        cout<<endl;
    }
    return 0;
}