1121 Damn Single (25)

"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 (<=50000), 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 (<=10000) 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

题目大意：给出n对数，每一对数表示一对情侣的id；再给出m个id进行查询，找出没有情侣，或者情侣没有出现的人，按顺序输出这些人的id。
思路：1.用数组来记录情侣关系cp[i]=j,cp[j]=i 表示i,j是情侣关系，不存在情侣的记为cp[i]=-1。 
     2.记录每一个人的出现与否，exist[i]=1，表示i出现，exis[i]=0 表示i没有出现，将出现的人依次记录在present[i]中
　　　3.满足条件的为没有情侣的--cp[present[i]]=-1,以及情侣没有出现的--exist[cp[present[i]]]==0; 
要求按id顺序输出，用set来保存满足条件的id，插入过程中，自动排序，也可以保存在数组中，最后再排序
注意点：当 输入项比较多的时候，应该避免使用cin,cout这两个的效率较低

#include<iostream>
#include<vector>
#include<set>
using namespace std;
int main(){
  vector<int> cp(100001,-1), present(100001), exist(100001,0);
  set<int> ans;
  int n, m, i, a, b, cnt=0;
  cin>>n;
  for(i=0; i<n; i++){
    scanf("%d%d",&a, &b);
    cp[a] = b;
    cp[b] = a;
  }
  cin>>m;
  for(i=0; i<m; i++){
    scanf("%d", &present[i]);
    exist[present[i]] = 1;
  }
  for(i=0; i<m; i++){
    if(!(cp[present[i]]!=-1 && exist[cp[present[i]]])){
      cnt++;
      ans.insert(present[i]);
    }
  }
  cout<<cnt<<endl;
  for(auto it=ans.begin(); it!=ans.end(); it++){
    if(it==ans.begin())  printf("%05d", *it);
    else  printf(" %05d", *it);
  }
}