"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
使用couple记录是否有配偶
使用flag标记是否配偶到场
注意：最后不要输出换行，不然一旦没有输出，那么就有两个换行了！
1 #include <iostream> 2 #include <set> 3 using namespace std; 4 int main() 5 { 6 int n, m, boy, girl, id; 7 int couple[100005], flag[100005]; 8 set<int>guests, res; 9 cin >> n; 10 fill(couple, couple + 100005, -1); 11 fill(flag, flag + 100005, 0); 12 while (n--) 13 { 14 cin >> boy >> girl; 15 couple[boy] = girl; 16 couple[girl] = boy; 17 } 18 cin >> m; 19 while (m--) 20 { 21 cin >> id; 22 guests.insert(id); 23 if (couple[id] == -1) 24 continue; 25 else if (flag[id] == 0)//对偶没有来 26 { 27 flag[id] = 1; 28 flag[couple[id]] = 1; 29 } 30 else if (flag[id] == 1)//对偶来了 31 { 32 flag[id] = -1; 33 flag[couple[id]] = -1; 34 } 35 } 36 for (auto a : guests) 37 { 38 if (flag[a] != -1) 39 res.insert(a); 40 } 41 cout << res.size() << endl; 42 for (auto a : res) 43 printf("%s%05d", (a == *(res.begin()) ? "" : " "), a); 44 return 0; 45 }