2015 HUAS Summer Training#1~D

10763 Foreign Exchange
Your non-profit organization (iCORE - international Confederation of Revolver Enthusiasts) coordinates a very successful foreign student exchange program. Over the last few years, demand has sky-rocketed and now you need assistance with your task.
The program your organization runs works as follows: All candidates are asked for their original location and the location they would like to go to. The program works out only if every student has a suitable exchange partner. In other words, if a student wants to go from A to B, there must be another student who wants to go from B to A. This was an easy task when there were only about 50 candidates, however now there are up to 500000 candidates!
Input
The input file contains multiple cases. Each test case will consist of a line containing n – the number of candidates (1 ≤ n ≤ 500000), followed by n lines representing the exchange information for each candidate. Each of these lines will contain 2 integers, separated by a single space, representing the candidate’s original location and the candidate’s target location respectively. Locations will be represented by nonnegative integer numbers. You may assume that no candidate will have his or her original location being the same as his or her target location as this would fall into the domestic exchange program. The input is terminated by a case where n = 0; this case should not be processed.
Output
For each test case, print ‘YES’ on a single line if there is a way for the exchange program to work out, otherwise print ‘NO’.
Sample Input
십
1 2
2 1
3 4
4 3
100 200
200 100
57 2
2 57
1 2
2 1
십
1 2
3 4
5 6
7 8
9 10
11 12
13 14
15 16
17 18
19 20
0
Sample Output
YES
NO
문제 풀이 사고방식: 우선 이 문제를 보면 정렬 문제일 것이다.이것도 0을 끝 표시로 하기 때문에 입력할 때마다 입력한 숫자가 0인지 아닌지를 먼저 판단해야 한다.제목은 교환학생이 깊이 연구하고 공부하는 것을 의미한다. 첫 번째 숫자는 이 학생의 현재 소재지를 나타내고, 두 번째 숫자는 이 학생의 목표 장소를 나타낸다.사실 간단하게 말하면 첫 번째 열과 두 번째 열을 각각 정렬하고 두 열의 정렬 결과가 같은지 아닌지를 보는 것이다. 만약 그렇다면'YES'를 출력하고, 그렇지 않으면'NO'를 출력한다.
프로그램 코드:

#include<cstdio> #include<algorithm> using namespace std; const int maxn=500000; int a[maxn],b[maxn]; int main() { int n,count; while(scanf("%d",&n)==1&&n) { count=0; for(int i=0;i<n;i++) scanf("%d%d",&a[i],&b[i]); sort(a,a+n); sort(b,b+n); for(int j=0;j<n;j++) if(a[j]==b[j]) count++; if(count==n) printf("YES
"); else printf("NO
"); } return 0; }