"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
注意00000的情况,我wa了一次就是因为这个。
1 #include <bits/stdc++.h> 2 using namespace std; 3 int n,m; 4 map<string,string> mp; 5 string an[100000]; 6 map<string, int> vis; 7 vector<string> v; 8 int main(){ 9 cin >> n; 10 string x,y; 11 for(int i = 0 ; i < n; i++){ 12 cin >> x >> y; 13 mp[x] = y; 14 mp[y] = x; 15 } 16 cin >> m; 17 for(int i = 0 ; i < m; i++){ 18 cin >> an[i]; 19 vis[an[i]] = 1; 20 } 21 for(int i = 0; i < m; i++){ 22 string ans = mp[an[i]]; 23 if(vis[ans] != 1){ 24 v.push_back(an[i]); 25 } 26 } 27 sort(v.begin(), v.end()); 28 cout << v.size()<<endl; 29 for(int i = 0 ; i < v.size(); i++){ 30 printf("%s%c", v[i].c_str(), i == v.size()-1?' ':' '); 31 } 32 return 0; 33 }