第一次用了两重循环，结果TLE了。然后把每个学生对应到一个映射里，如果有意愿相同的那么数目加一，最后比较意愿相对的人数就OK了。说的有点乱，不过想出来了就很顺了。



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

struct Ex_student {
    int x, y;
};

struct Ex_cmp {
    bool operator () (const Ex_student &k1, const Ex_student &k2) const {
        if(k1.x != k2.x)
            return k1.x < k2.x;
        if(k1.y != k2.y)
            return k1.y < k2.y;
        return false;
    }
};

int main() {
    int n;
    while (cin >> n&& n) {
        bool exsit = true;
        vector<Ex_student> s;
        map<Ex_student, int, Ex_cmp> student_list;

        for(int i = 0; i < n; i++) {
            Ex_student student;
            cin >> student.x >> student.y;
            if(student_list.count(student))
                student_list[student]++;
            else {
                student_list[student] = 1;
            }
            s.push_back(student);
        }

        for(int i = 0; i < n; i++) {
            Ex_student student;
            student.x = s[i].y;
            student.y = s[i].x;
            if(student_list.count(student)&&
               student_list[student] == student_list[s[i]])
               ; else {
                    exsit = false;
                    break;
                }
        }
        exsit ? cout << "YES" << endl
        : cout << "NO" << endl;
    }
    return 0;
}