方法1

#include <stdio.h>
#define BLANK -1
#define SIGNED -2
#define SINGLE -3

int main()
{
    int couple[100000], count = 0, N, M, ID1, ID2;
    for(int i = 0; i < 100000; i++) 
        couple[i] = BLANK;
    
    /* Read 'couple-list', every pair of 'index' and 'value' are a couple. */
    scanf("%d", &N);
    for(int i = 0; i < N; i++)
    {
        scanf("%d %d", &ID1, &ID2);
        couple[ID1] = ID2;
        couple[ID2] = ID1;
    }
    
    scanf("%d", &M);
    for(int i = 0; i < M; i++)          /* Read guest list. */
    {
        scanf("%d", &ID1);
        if(couple[ID1] >= 0)                /* If one has a mate */
            couple[ID1] = SIGNED;           /*   set SIGNED */
        else                                /* Else: not in the 'couple-list' */
            couple[ID1] = SINGLE, count++;  /*   set SINGLE */
    }
    
    /* If couple[ID] is >= 0 (but not signed) but couple[couple[ID]] == SIGNED
     * (signed in), this means 'ID' didn't come but his/her mate did.
     * So his/her mate is alone in the party, set to SINGLE. */
    for(int i = 0; i < 100000; i++) 
        if(couple[i] >= 0 && couple[couple[i]] == SIGNED) 
            couple[couple[i]] = SINGLE, count++;

    /* Those whose value is SINGLE is either a bachelor or came alone */
    printf("%d\n", count);
    for(int i = 0; i < 100000; i++) 
        if(couple[i] == SINGLE)
            printf("%05d%c", i, --count ? ' ' : '\0');
    
    return 0;
}

方法2

#include<iostream>
#include<map>
#include<string>
using namespace std;
int main(){
    map<string, pair<string,int>> s, res;     //s用于判断是否单身
    map<string, pair<string,int>>::iterator it;
    int n, m, count = 0;
    string r, l;
    cin >> n;
    while (n--) {
        cin >> l >> r;
        s[l].first = r;     //存入伴侣信息
        s[r].first = l;
    }
    cin >> m;
    for (int i = 0; i < m; i++) {
        cin >> l;
        s[l].second++;
    }
    for (it = s.begin(); it != s.end(); it++) {
        if (it->second.second && !s[it->second.first].second) {  //逻辑判断：自己出席了会议，伴侣没有
            count++;               //记录单身狗数目
            res[it->first].second++;    //将符合条件的人存到res中
        }
    }
    cout << count << endl;
    if (count) {        //注意count为0时，就不用输出ID了，否则测试点2出现段错误、运行超时
        it = res.begin();
        cout << it++->first;
        for (; it != res.end(); it++) {
            cout << " " << it->first;//这样能输出空格
        }
    }
    return 0;
}