1121 Damn Single (25 point(s))

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本文介绍了一个有趣的算法问题，即在已知情侣配对的情况下，如何找出参加派对的单身人士。通过建立哈希表并遍历所有在场嘉宾，算法能够高效地识别并输出单身者名单。

1121 Damn Single (25 point(s))

"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

哈希、STL。

首先建立哈希表，假设每一个在场的人都是单身；然后遍历所有的夫妻数对，如果它们都在场，则说明他们不会孤独，于是取消标记、在场单身人数减2。

#include<iostream>
#include<cstring>
#include<vector> 
using namespace std;
vector<pair<int,int> > v;
const int MAX = 1e5+7;
bool mark[MAX]={false};
int main(void){
	int N,M,a,b;cin>>N;
	while(N--){
		cin>>a>>b;
		v.push_back(pair<int,int>(a,b));
	}
	cin>>M;
	for(int i=0;i<M;i++){
		cin>>a;mark[a]=true;
	}
	for(auto i :v){
		if(mark[i.first]&&mark[i.second]){
			M-=2;
			mark[i.first]=false;
			mark[i.second]=false;
		}
	}
	bool flag = false;cout<<M<<endl;
	for(int i=0;i<100001;i++){
		if(mark[i]){
			if(!flag) printf("%05d",i),flag=true;
			else printf(" %05d",i);
		}
	}
	return 0;
}