1121 Damn Single

"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

题意：

　　找出查询序列中的单身人数。

思路：

　　用set将party上所有的人都存储起来，然后查找是否有couples在set中如果有的话则将其删除。

Code：

#include <bits/stdc++.h>

using namespace std;

int main() {
    int n, m;
    cin >> n;
    vector<pair<int, int> > v;
    int p1, p2;
    for (int i = 0; i < n; ++i) {
        cin >> p1 >> p2;
        v.push_back({p1, p2});
    }
    cin >> m;
    int p;
    set<int> s;
    while (m--) {
        cin >> p;
        s.insert(p);
    }
    for (int i = 0; i < n; ++i) {
        auto it1 = s.find(v[i].first);
        auto it2 = s.find(v[i].second);
        if (it1 != s.end() && it2 != s.end()) {
            s.erase(it1);
            s.erase(it2);
        }
    }
    cout << s.size() << endl;
    bool isFirst = true;
    for (int i : s) {
        if (isFirst) {
            printf("%04d", i);
            isFirst = false;
        } else {
            printf(" %04d", i);
        }
    }
    return 0;
}