제출 #164179

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
164179		dimash241	Zamjena (COCI18_zamjena)	C++17	70 / 70	236 ms	16988 KiB

zamjena

#pragma GCC target("avx2")
#pragma GCC optimize("O3")

# include <x86intrin.h>
# include <bits/stdc++.h>

# include <ext/pb_ds/assoc_container.hpp>
# include <ext/pb_ds/tree_policy.hpp>

using namespace __gnu_pbds;
using namespace std;
 
template<typename T> using ordered_set = tree <T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

#define _USE_MATH_DEFINES_
#define ll long long
#define ld long double
#define Accepted 0
#define pb push_back
#define mp make_pair
#define sz(x) (int)(x.size())
#define every(x) x.begin(),x.end()
#define F first
#define S second
#define lb lower_bound
#define ub upper_bound
#define For(i,x,y)  for (ll i = x; i <= y; i ++) 
#define FOr(i,x,y)  for (ll i = x; i >= y; i --)
#define SpeedForce ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0)
// ROAD to...                                                                                                                                                                                                                Red

inline void Input_Output () {
	//freopen(".in", "r", stdin);
   //freopen(".out", "w", stdout);
}

const double eps = 0.000001;
const ld pi = acos(-1);
const int maxn = 1e7 + 9;
const int mod = 1e9 + 7;
const ll MOD = 1e18 + 9;
const ll INF = 1e18 + 123;
const int inf = 2e9 + 11;
const int mxn = 1e6 + 9;
const int N = 6e4 + 123;                                          
const int M = 22;
const int pri = 997;
const int Magic = 2101;

const int dx[] = {-1, 0, 1, 0};
const int dy[] = {0, -1, 0, 1};
 
int n, m, k;
map < string, vector < string > > g;
map < string, bool > u;
string a[N];
string b[N];

int dfs (string s) {
	u[s] = 1;
	int res = 0;
	if (isdigit(s[0])) res++;
	for (auto to : g[s]) if (!u.count(to)) {
		res += dfs(to);
	}
	return res;
}

int main () {
   SpeedForce;

   cin >> n;

   For(i, 1, n) cin >> a[i];
   For(i, 1, n) cin >> b[i];

   For (i, 1, n) if (a[i] != b[i]) {
   	g[a[i]].pb(b[i]);
   	g[b[i]].pb(a[i]);
   }

   For (i, 1, n) if (a[i] != b[i] && !u[a[i]] && dfs(a[i]) > 1) {
   	cout << "NE\n";
   	exit(0);
   }

   cout << "DA\n";
	
	return Accepted;
}

// B...a

• Subtask 1: 14 / 14
• Subtask 2: 14 / 14
• Subtask 3: 14 / 14
• Subtask 4: 14 / 14
• Subtask 5: 14 / 14