1121 Damn Single （25 分）

"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

解析：該題的關鍵點在於一個夫婦中只出現一人和兩人都出現如何區分。這裡用isExist去判斷。例如夫婦中的男方出現，不知道男方是否能輸出，但女方一定要標記為不能輸出。這樣處理即可輕鬆AC。

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

int main(){
	int n,m,a,b;
	cin>>n;
	vector<int> couple( 100001,-1);
	for(int i=1;i<=n;i++){
		cin>>a>>b;
		couple[a] = b;
		couple[b] = a;
	}
	cin>>m;
	vector<int> vec(m),isExist(100001,0);
	for(int i=0;i<m;i++){
		cin>>vec[i];
		if(couple[vec[i]] != -1){
			//夫婦中的男方出現，不知道男方是否能輸出，但女方一定不能輸出 
			isExist[couple[vec[i]]] = 1;
		}			
	}
	set<int> s;
	for(int i=0;i<m;i++)
		if(isExist[vec[i]] == 0)
			s.insert(vec[i]);
	
	int cnt = s.size(),i = 0;
	cout<<cnt<<endl;
	for(auto it = s.begin();it!=s.end();it++)
		printf("%s%05d",it!=s.begin()?" ":"",*it);
	//for(auto it:s){
	//	printf("%05d%s",it,i!=cnt - 1?" ":"");
	//	i++;
	//}
		
	return 0;
}