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

标记每一对情侣，然后查看有情侣的，情侣是否在场就好。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int love[],haslove[],loveex[],a,b,c;
int ans[],peo[],n,m;
int main()
{
    cin>>n;
    for(int i = ;i < n;i ++)
    {
        cin>>a>>b;
        love[a] = b;
        love[b] = a;
        haslove[a] = ;
        haslove[b] = ;
    }
    cin>>m;
    for(int i = ;i < m;i ++)
    {
        cin>>peo[i];
        if(haslove[peo[i]])loveex[love[peo[i]]] = ;
    }
    for(int i = ;i < m;i ++)
    {
        if(!loveex[peo[i]])ans[c ++] = peo[i];
    }
    sort(ans,ans + c);
    cout<<c<<endl;
    if(c)printf("%05d",ans[]);
    for(int i = ;i < c;i ++)
        printf(" %05d",ans[i]);
}