A1121. Damn Single

"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

#include<cstdio>
#include<iostream>
#include<vector>
using namespace std;
int couple[100000], guest[100000] = {0,0};
int main(){
    int N, M;
    fill(couple, couple + 100000, -1);
    scanf("%d", &N);
    for(int i = 0; i < N; i++){
        int cp1,cp2;
        scanf("%d%d", &cp1, &cp2);
        couple[cp1] = cp2;
        couple[cp2] = cp1;
    }
    scanf("%d", &M);
    for(int i = 0; i < M; i++){
        int cp;
        scanf("%d", &cp);
        guest[cp] = 1;
    }
    vector<int> ans;
    for(int i = 0; i < 100000; i++){
        if(guest[i] == 1 && (couple[i] == -1 || guest[couple[i]] == 0))
            ans.push_back(i);
    }
    printf("%d\n", ans.size());
    for(int i = 0; i < ans.size(); i++){
        if(i == ans.size() - 1)
            printf("%05d", ans[i]);
        else printf("%05d ", ans[i]);
    }
    cin >> N;
    return 0;
}