PAT 1121 Damn Single[简单]

1121 Damn Single （25 分）

"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

 题目大意：给出n对情侣，然后再给出m对参加宴会的人，判断这些人当中有多少人是没有伙伴来参加宴会的（本身已经结婚，但是没带伴侣来的也会被计算进来）

#include <iostream>
#include<vector>
#include<map>
using namespace std;

map<string,string> m2f;
map<string,string> inp;
int main() {
    int n;
    cin>>n;
    string x,y;
    for(int i=0;i<n;i++){
        cin>>x>>y;
        m2f[x]=y;
        m2f[y]=x;
    }
    int m;
    cin>>m;
    string s;
    for(int i=0;i<m;i++){
        cin>>s;
        inp[s]=1;//因为这里的map是自然排序的，所以最终vector里也是自然排序的。
    }
    int ct=0;
    vector<string> vt;
    for(auto it=inp.begin();it!=inp.end();it++){
        string str=it->first;
        if(inp.count(m2f[str])==0){
            vt.push_back(str);
        }
    }
    cout<<vt.size()<<'\n';
    for(int i=0;i<vt.size();i++){
        cout<<vt[i];
        if(i!=vt.size()-1)cout<<" ";
    }
    return 0;
}

//还是比较简单的，但是还是提交了两次，为什么呢？

1.需要主要最后的输出格式，是1!=vt.size()-1，知道了吗？