"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
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1 #include<cstdio> 2 #include<iostream> 3 #include<vector> 4 using namespace std; 5 int couple[100000], guest[100000] = {0,0}; 6 int main(){ 7 int N, M; 8 fill(couple, couple + 100000, -1); 9 scanf("%d", &N); 10 for(int i = 0; i < N; i++){ 11 int cp1,cp2; 12 scanf("%d%d", &cp1, &cp2); 13 couple[cp1] = cp2; 14 couple[cp2] = cp1; 15 } 16 scanf("%d", &M); 17 for(int i = 0; i < M; i++){ 18 int cp; 19 scanf("%d", &cp); 20 guest[cp] = 1; 21 } 22 vector<int> ans; 23 for(int i = 0; i < 100000; i++){ 24 if(guest[i] == 1 && (couple[i] == -1 || guest[couple[i]] == 0)) 25 ans.push_back(i); 26 } 27 printf("%d ", ans.size()); 28 for(int i = 0; i < ans.size(); i++){ 29 if(i == ans.size() - 1) 30 printf("%05d", ans[i]); 31 else printf("%05d ", ans[i]); 32 } 33 cin >> N; 34 return 0; 35 }