【笨方法学PAT】1121 Damn Single （25 分）

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一、题目

"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

二、题目大意

跟男女朋友名单和出席的人，判断单身狗和只有一个人的人的名单。

三、考点

数组

四、注意

1、使用数组保存情侣关系；

2、单身或者TA没有参加的都列为单身狗。

五、代码

#include<iostream>
#include<map>
#include<set>
#include<vector>
#define N 100001
using namespace std;
int n1[N] = { -1 };
bool show[N] = { false };
int main() {
	//read
	int n;
	cin >> n;
	for (int i = 0; i < n; ++i) {
		int a, b;
		cin >> a >> b;
		n1[a] = b;
		n1[b] = a;
	}

	int m;
	cin >> m;
	vector<int> v(m);
	set<int> sset;
	for (int i = 0; i < m; ++i) {
		cin >> v[i];
		show[v[i]] = true;
	}

	//solve
	for (int i = 0; i < m; ++i) {
		//damn
		if (n1[v[i]] == -1) {
			sset.insert(v[i]);
			continue;
		}

		//just one
		if(show[n1[v[i]]]==false) {
			sset.insert(v[i]);
			continue;
		}
	}

	//output
	cout << sset.size() << endl;
	for (auto it = sset.begin(); it != sset.end(); ++it) {
		if (it != sset.begin())
			cout << " ";
		printf("%05d", *it);
	}

	system("pause");
	return 0;
	
}