PAT 1121 Damn Single

本文介绍了一个用于识别参加聚会中单身人士的算法。该算法通过输入已知情侣的ID来找出那些独自出席的人,并确保每人的ID唯一对应。文章提供了完整的C++实现代码,包括输入输出示例。

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1121. Damn Single (25)

"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 (<=50000), 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 (<=10000) 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
#include <cstdio>
#include <cstring>
#include <vector>
#include <map>
#include <algorithm>
#include <iostream>
using namespace std;
int main(){
	int n,count=0;
	string a,b;
	map<string,string> c;
	map<string,int> d;
	vector<string> re;
	cin>>n;
	for(int i=1;i<=n;i++){
		cin>>a>>b;
		c.insert(pair<string,string>(a,b));
		c.insert(pair<string,string>(b,a));
	}
	cin>>n;
	for(int i=1;i<=n;i++){
		cin>>a;
		d[a]=1;
	}
	int flag=0;
	map<string,int> ::iterator it;
	for(it=d.begin();it!=d.end();it++){
		if(d.count(c[it->first])==0){
			count++;
			re.push_back(it->first);
		}
	}
	cout<<count<<endl;
	for(int i=0;i<re.size();i++){
		cout<<re.at(i)<<(i==re.size()-1?'\n':' ');
	}
	return 0;
}


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