PAT Class B-Single Dog

The title description
"Single Dog" is a Chinese nickname for single people.

This question asks you to find out the single guests from a large party of tens of thousands of people, so that you can give special care.

Input format The
first line of input gives a positive integer N, which is the logarithm of the known couple/partner; the
subsequent N lines, each line gives a couple/partner-for convenience, each person corresponds to an ID number, which is 5 digits (from 00000 to 99999), the IDs are separated by spaces;
after that, a positive integer M is given, which is the total number of people participating in the party; the
next line gives the IDs of the M guests, separated by spaces.

The title guarantees that no one has bigamy or two boats.

Output format
First, the first line outputs the total number of guests who placed orders;
then the second line lists the guests who placed orders in ascending order of ID. The IDs are separated by a space, and there must be no extra spaces at the beginning and end of the line.

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

Output sample
5
10000 23333 44444 55555 88888

Data range
N ≤ 50 000
M ≤ 10 000

---

Problem solution
STL:

解题步骤：

1. First with mapeach mutually couple mapping;
2. And then setrecorded to the party of people;
3. Finally enumerate all setthe elements, if their spouse is not set, then added to ansgo;

#include <iostream>
#include <unordered_map>
#include <vector>
#include <set>
using namespace std;

int n, m;
vector<string> ans;
unordered_map<string, string> H;
set<string> S;

int main()
{
    
    
	cin >> n;
	
	for (int i = 1; i <= n; i ++)
	{
    
    
		string a, b;
		cin >> a >> b;
		H[a] = b, H[b] = a;
	}
	
	cin >> m;
	
	for (int i = 1; i <= m; i ++) 
	{
    
    
		string s;
		cin >> s;
		S.insert(s);
	}
	
	for (auto &x : S)
		if(!S.count(H[x])) ans.push_back(x);
		
	cout << ans.size() << endl;
	
	bool flag = true;
	for (int i = 0; i < ans.size(); i ++)
		if(flag) cout << ans[i], flag = false;
		else cout << ' ' << ans[i];
	
	return 0;		
}

Misunderstanding: Test point 2 segment error, but no error can be found

#include <iostream>
#include <map>
#include <set>
using namespace std;

int n, m;
map<string, string> H;
set<string> S;

int main()
{
    
    
	cin >> n;
	
	for (int i = 1; i <= n; i ++)
	{
    
    
		string a, b;
		cin >> a >> b;
		H[a] = b, H[b] = a;
	}
	
	cin >> m;
	
	for (int i = 1; i <= m; i ++) 
	{
    
    
		string s;
		cin >> s;
		S.insert(s);
	}
	
	for (auto &x : S)
		if(S.count(H[x])) 
		{
    
    
			S.erase(S.find(x));
			S.erase(S.find(H[x]));
			if(S.empty()) break;
		}
		
	cout << S.size() << endl;
	
	bool flag = true;
	for (auto &x : S)
		if(flag) cout << x, flag = false;
		else cout << ' ' << x;
	
	return 0;		
}