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
10
1 2
2 1
3 4
4 3
100 200
200 100
57 2
2 57
1 2
2 1
10
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
题目类型：二分枚举
算法分析：由于数据组数达到5e5，所以考虑使用二分枚举的方法。使用map<pair<int, int>, int>存储每个对应关系的组数，最后枚举一遍若关系(a,b)的组数和(b,a)的组数不同，则不可能交换
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/************************************************** filename :d.cpp author :maksyuki created time :2018/1/22 20:41:01 last modified :2018/1/22 20:51:16 file location :C:\Users\abcd\Desktop\TheEternalPoet ***************************************************/ #pragma comment(linker, "/STACK:102400000,102400000") #include <set> #include <bitset> #include <list> #include <map> #include <stack> #include <queue> #include <deque> #include <string> #include <vector> #include <ios> #include <iostream> #include <fstream> #include <sstream> #include <iomanip> #include <algorithm> #include <utility> #include <complex> #include <numeric> #include <functional> #include <cmath> #include <ctime> #include <climits> #include <cstdarg> #include <cstdio> #include <cstdlib> #include <cstring> #include <cctype> #include <cassert> using namespace std; #define CFF freopen ("in", "r", stdin) #define CFO freopen ("out", "w", stdout) #define CPPFF ifstream cin ("in") #define CPPFO ofstream cout ("out") #define DB(ccc) cout << #ccc << " = " << ccc << endl #define DBT printf("time used: %.2lfs\n", (double) clock() / CLOCKS_PER_SEC) #define PB push_back #define MP(A, B) make_pair(A, B) typedef long long LL; typedef unsigned long long ULL; typedef double DB; typedef pair <int, int> PII; typedef pair <int, bool> PIB; const int INF = 0x7F7F7F7F; const int MOD = 1e9 + 7; const double EPS = 1e-10; const double PI = 2 * acos (0.0); const int maxn = 1e5 + 6666; map<PII, int> ma; int main() { #ifdef LOCAL CFF; //CFO; #endif int n; while(scanf("%d", &n) != EOF) { if(!n) break; ma.clear(); int va, vb; for(int i = 1; i <= n; i++) { scanf("%d %d", &va, &vb); if(!ma.count(PII(va, vb))) ma[PII(va,vb)] = 1; else ma[PII(va,vb)]++; } bool ok = true; for(map<PII, int>:: iterator it = ma.begin(); it != ma.end(); it++) { int ta = (it->first).first, tb = (it->first).second, tcnt = it->second; if(ma.count(PII(tb,ta)) && ma[PII(tb,ta)] == tcnt) continue; else { ok = false; break; } } if(ok) puts("YES"); else puts("NO"); } return 0; } |
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