"Damn Single (單身狗)" is the Chinese nickname for someone who is being single. You are supposed to find those who are alone in a big party, so they can be taken care of.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (≤ 50,000), the total number of couples. Then N lines of the couples follow, each gives a couple of ID‘s which are 5-digit numbers (i.e. from 00000 to 99999). After the list of couples, there is a positive integer M (≤ 10,000) followed by M ID‘s of the party guests. The numbers are separated by spaces. It is guaranteed that nobody is having bigamous marriage (重婚) or dangling with more than one companion.

Output Specification:

First print in a line the total number of lonely guests. Then in the next line, print their ID‘s in increasing order. The numbers must be separated by exactly 1 space, and there must be no extra space at the end of the line.

Sample Input:

3
11111 22222
33333 44444
55555 66666
7
55555 44444 10000 88888 22222 11111 23333
 

Sample Output:

5
10000 23333 44444 55555 88888


題意：找出誰是“單身狗”。。。先輸入各對CP，然後再輸入參加party的人員，統計輸入的這些人裏哪些是沒有CP一起來的，升序輸出。

分析：水題。。我思路是用一個數組couple，初始化為-1（因為人的編號為00000-99999），作用是來存儲某人的CP，如11111的CP是22222，那麽couple[11111]=22222,couple[22222]=11111。最後因為要格式輸出，不知道叠代器怎麽格式輸出。。就先把結果存到了vector中，具體見代碼

 1 /**
 2 * Copyright(c)
 
 3 * All rights reserved.
 4 * Author : Mered1th
 5 * Date : 2019-02-27-21.51.47
 6 * Description : A1121
 7 */
 8 #include<cstdio>
 9 #include<cstring>
10 #include<iostream>
11 #include<cmath>
12 #include<algorithm>
13 #include<string>
14 #include<unordered_set>
15 #include<map>
16 #include<vector>
17 #include<set>
18 using namespace std;
19 const int maxn=100010;
20 int couple[maxn]={-1};
21 int main(){
22 #ifdef ONLINE_JUDGE
23 #else
24     freopen("1.txt", "r", stdin);
25 #endif
26     int n,m,u,v,t;
27     scanf("%d",&n);
28     for(int i=0;i<n;i++){
29         scanf("%d%d",&u,&v);
30         couple[u]=v;
31         couple[v]=u;
32     }
33     scanf("%d",&m);
34     vector<int> tem;
35     for(int i=0;i<m;i++){
36         scanf("%d",&t);
37         tem.push_back(t);
38     }
39     set<int> ans;
40     for(int i=0;i<m;i++){
41         int coup=couple[tem[i]];
42         if(coup==-1) ans.insert(tem[i]);
43         else{
44             if(find(tem.begin(),tem.end(),coup)==tem.end()){
45                 ans.insert(tem[i]);
46             }
47         }
48     }
49     cout<<ans.size()<<endl;
50     vector<int> res;
51     for(auto it=ans.begin();it!=ans.end();it++){
52         res.push_back(*it);
53     }
54     for(int i=0;i<res.size();i++){
55         printf("%05d",res[i]);
56         if(i!=res.size()-1) printf(" ");
57     }
58     return 0;
59 }

1121 Damn Single （25 分）