UVa 10763
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++;
else {
student_list = 1;
}
s.push_back(student);
}
for(int i = 0; i < n; i++) {
Ex_student student;
student.x = s.y;
student.y = s.x;
if(student_list.count(student)&&
student_list == student_list])
; else {
exsit = false;
break;
}
}
exsit ? cout << "YES" << endl
: cout << "NO" << endl;
}
return 0;
}
页:
[1]