1121. Damn Single (25) map,set

"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:

311111 2222233333 4444455555 66666755555 44444 10000 88888 22222 11111 23333

Sample Output:

510000 23333 44444 55555 88888

这题我记得当时考PAT乙的最后一题。。当时还不知道set，map。写的特别复杂。很多嵌套的循环（代码没保存，当时AC了，后面就没做）

用map建立关系，如果map【a】!=0，而且map[a]不存在set中，就是单身狗

或者说map【a】==0，肯定是单身狗

最后保存输出

#include<cstdio>#include<cmath>#include<algorithm>#include<iostream>#include<cstring>#include<queue>#include<vector>#include<set>#include<map>#include<stack>using namespace std;int main(){int n;cin>>n;map<int,int> m;for(int i=0;i<n;i++){    int a,b;    cin>>a>>b;    m[a+1]=b+1;    m[b+1]=a+1;}int k;cin>>k;set<int> s;for(int i=0;i<k;i++){int num;cin>>num;num++;s.insert(num);} int flag=1;int out[100001],cnt=0; for(set<int>::iterator it=s.begin();it!=s.end();it++){int num=*it;if(m[num]==0){ out[cnt++]=num-1;}else{    if(s.find(m[num])!=s.end()){    continue;}else{out[cnt++]=num-1;}}} cout<<cnt<<endl;for(int i=0;i<cnt;i++){if(i==0) printf("%05d",out[i]);else printf(" %05d",out[i]);}return 0;}