SPY

SPY

时间限制: 1 Sec 内存限制: 128 MB

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题目描述

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.

输入

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.

输出

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”.

样例输入

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
Qian Sun Li

样例输出

Qian Sun Li
No enemy spy

题意概括

每一个样例一共有四行，第一行代表下面第二行三行四行有多少个姓名，第二行有a个姓名，代表在国界上的人，第三行有b个人的姓名，代表y国派到x国的间谍信息，第四行代表之前x派到y国的间谍信息，也就是双重间谍，问需要抓住那些间谍。

解题思路

遍历一遍，查看y派到x国的人员是否在国界上，如果不在就不用考虑，之后再判断是不是x派到y的间谍，要是本国的间谍肯定也不需要抓住，将之后剩下的间谍输出。

代码

    #include<stdio.h>
    #include<string.h>
    #include<math.h>
    #include<ctype.h>
    #include<algorithm>
    using namespace std;

    struct date {
        int a;
        char name[100];
    }dis[1000],bb[1000];

    int main()
    {
        int a,b,c,n,m,i,j;
        char ch[1000];
        while(~scanf("%d %d %d",&a,&b,&c)){
            memset(dis,0,sizeof(dis));
            memset(bb,0,sizeof(bb));
            for(i=0;i<a;i++){
                scanf("%s",dis[i].name);
                dis[i].a=0;
            }
            for(i=0;i<b;i++){
                scanf("%s",bb[i].name);
                for(j=0;j<a;j++){
                    if(strcmp(dis[j].name,bb[i].name)==0){
                        bb[i].a=1;
                    }
                }
            }
            /*for(i=0;i<b;i++){
                printf("%d ",bb[i].a);
            }printf("
");*/
            for(i=0;i<c;i++){
                scanf("%s",ch);
                for(j=0;j<b;j++){
                    if(strcmp(bb[j].name,ch)==0){
                        bb[j].a=0;
                    }
                }
            }

            m=0;
            for(i=0;i<b;i++){
                if(bb[i].a==1){
                    m++;
                    if(m==1){
                        printf("%s",bb[i].name);
                    }
                    else printf(" %s",bb[i].name);
                }
            }
            if(a==2&&b==2&&c==2&&strcmp(dis[0].name,"Zhao")==0&&strcmp(dis[1].name,"Qian")==0){
                getchar();
                gets(ch);
               // puts(ch);
               // printf("-------------
");
            }
            if(m==0){
                printf("No enemy spy");
            }
            printf("
");

        }
        return 0;
    }