PAT A1121 Damn Single （25 分）——set遍历

"Damn Single (单身狗)" 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

 #include <stdio.h>
 #include <algorithm>
 #include <queue>
 #include <vector>
 #include <set>
 #include <map>
 using namespace std;
 const int maxn=;
 int couple[maxn]={};
 vector<int> v;
 int peop[maxn]={};
 int n;
 int main(){
     scanf("%d",&n);
     for(int i=;i<n;i++){
         int cp,cp2;
         scanf("%d %d",&cp,&cp2);
         couple[cp]=cp2;
         couple[cp2]=cp;
     }
     scanf("%d",&n);
     for(int i=;i<n;i++){
         int p;
         scanf("%d",&p);
         peop[p]=;
         v.push_back(p);
     }
     for(int i=;i<n;i++){
         if(peop[v[i]]==){
             if(peop[couple[v[i]]]== && couple[couple[v[i]]]==v[i]){
                 peop[v[i]]=;
                 peop[couple[v[i]]]=;
             }
         }
     }
     int cnt=;
     for(int i=;i<n;i++){
         if(peop[v[i]]==){
             cnt++;
         }
     }
     printf("%d\n",cnt);
     int num=;
     sort(v.begin(),v.end());
     for(int i=;i<n;i++){
         if(peop[v[i]]==){
             printf("%05d",v[i]);
             num++;
             if(num!=cnt)printf(" ");
         }
     }
 }

注意点：一开始想用set和map，发现map其实就是一个大数组，还不如直接开个大数组。一开始题目理解错了，以为只要找给的人里是单身狗的就行，结果是有对象的对象没来也算单身，行吧