1065 单身狗

题目传送门：https://pintia.cn/problem-sets/994805260223102976/problems/994805266942377984

题解：

#include<map>
#include<set>
#include<string>
#include<iomanip>
#include<iostream>
using namespace std;

int main() {
    set<int>s, ans;
    map<int, int>Wife;
    int n, num1, num2, M;
    cin >> n;
    while (n--) {
        cin >> num1 >> num2;
        Wife[num1] = num2;
        Wife[num2] = num1;
    }
    cin >> M;
    int *p = new int[M];
    for (int i = 0; i < M; i++) {
        cin >> p[i];
        s.insert(p[i]);
    }
    for (int i = 0; i < M; i++) {
        if (s.find(Wife[p[i]]) == s.end()) {//Wife[p[i]]代表其配偶，在集合中没找到
            ans.insert(p[i]);//放到新的集合中
        }
    }
    cout << ans.size() << endl;
    set<int>::iterator it;
    for (it = ans.begin(); it != ans.end(); it++) {
        if (it != ans.begin())
            cout << " ";
        cout << setw(5) << setfill('0') << *it;//注意不足5位补0
    }
    return 0;
}

 

转载于:https://www.cnblogs.com/Gzu_zb/p/10543001.html