"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
单身狗,我们给定一些夫妻,给定一些人员,把单身狗打印。(也要把我打印QAQ)
#include <iostream> #include <unordered_map> #include <set> using namespace std; int N, M, a, b; unordered_map<int, set<int>> s; set<int> query; int main() { scanf("%d", &N); while(N--) { scanf("%d%d", &a, &b); s[a].insert(b); s[b].insert(a); } scanf("%d", &M); while(M--) { scanf("%d", &a); query.insert(a); } for(auto& x: s) { auto it = query.find(x.first); if(it == query.end()) continue; for(auto& y: s[x.first]) { auto it2 = query.find(y); if(it2 != query.end()) { query.erase(it); query.erase(it2); break; } } } printf("%lu\n", query.size()); bool start = true; for(auto& x: query) { if(start == true) { printf("%05d", x); start = false; } else printf(" %05d", x); } return 0; }