题目描述：
"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
代码如下：
#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表示是否为夫妻，com表示是否来这次聚会（因为夫妻都来聚会才算）
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++) { //输入夫妻，并标记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) //如果没有对象，直接存入结果
res.push_back(vaule);
else {
s3[index++] = vaule; //否则存入中间数组，并标记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; //如果夫妻都来了，跳过这轮循环，否则存入结果
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;
}