PAT A1121 Damn Single (25point(s))

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"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

解答

#include<iostream>
#include<unordered_map>
#include<vector>
#include<algorithm>

using namespace std;

int main()
{
	int cn, gn, a, b;
	cin >> cn;
	unordered_map<int, int> mp;
	vector<int> ans;
	while (cn--) {
		cin >> a >> b;
		mp[a] = b;
		mp[b] = a;
	}
	cin >> gn;
	unordered_map<int, bool> guest;
	while (gn--) {
		cin >> a;
		guest[a] = true;
	}
	for (auto it = guest.begin(); it != guest.end(); it++) {
		if (mp.find(it->first) == mp.end() ||
			(mp.find(it->first) != mp.end() &&
				guest.find(mp[it->first]) == guest.end()))
			ans.push_back(it->first);
	}
	printf("%d\n", ans.size());
	if (ans.size() > 0) {
		sort(ans.begin(), ans.end());
		for (auto it = ans.begin(); it != ans.end() - 1; it++)
			printf("%05d ", *it);
		printf("%05d", ans.back());
	}
	return 0;
}