題目描述：
 

"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

程式碼如下：

#include <iostream>
#include <vector>
#include <algorithm>
#include <cstdlib>
#include <map>
#include <string>
using namespace std;
const int maxn = 50010;
int s3[maxn];
vector<int> res;

struct node {            //flag表示是否為夫妻，com表示是否來這次聚會（因為夫妻都來聚會才算）
	int num;
	int flag;
	int com;
	node() {
		num = 0;
		flag = 0;
		com = 0;
	}
};
map<int, node> mp;
int main()
{
	int s1, s2;
	int n, m, index = 0, vaule;
	cin >> n;
	for (int i = 0; i < n; i++) {        //輸入夫妻，並標記flag
		cin >> s1 >> s2;
		mp[s1].num = s2;
		mp[s2].num = s1;
		mp[s1].flag = 1;
		mp[s2].flag = 1;
	}
	cin >> m;
	for (int i = 0; i < m; i++) {        
		cin >> vaule;
		if (mp[vaule].flag == 0)        //如果沒有物件，直接存入結果
			res.push_back(vaule);
		else {
			s3[index++] = vaule;        //否則存入中間陣列，並標記com
			mp[vaule].com = 1;
		}
	}
	for (int i = 0; i < index ; i++) {
		if (mp[s3[i]].com == 1 && mp[mp[s3[i]].num].com == 1)
			continue;            //如果夫妻都來了，跳過這輪迴圈，否則存入結果
		else
			res.push_back(s3[i]);		
	}
	sort(res.begin(), res.end());
	printf("%d\n", res.size());
	for (int i = 0; i < res.size(); i++) {
		printf("%05d", res[i]);
		if (i < res.size() - 1)
			printf(" ");
	}
	system("pause");
	return 0;
}