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