Submission #251396

#	Time	Username	Problem	Language	Result	Execution time	Memory
251396		Hasti	Zamjena (COCI18_zamjena)	C++17	42 / 70	7 ms	3968 KiB

zamjena

///A drunk man will find his way home, but a drunk bird may get lost forever...
#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>;
typedef tree<double, null_type, less_equal<double>, rb_tree_tag, tree_order_statistics_node_update> ordered_set;
#define int long long
typedef long long ll;
typedef double dd;
typedef long double ldd;
typedef pair <int, int> pii;
typedef pair <pii , int> ppii;
#define pb push_back
#define pf push_front
#define ppb pop_back()
#define ppf pop_front()
#define f first
#define s second
#define sz(x) (int)x.size()
#define smin(x , y) x = min(x , y)
#define smax(x , y) x = max(x , y)
//#define mp make_pair
#define all(x) x.begin() , x.end()
#define debug(x) cerr<< #x << " = " << x << endl
#define ddebug(x, y) cerr<< #x << " = " << x  << ", " << #y << " = " << y << endl
#define uop(x, y) pii(min(x, y), max(x, y))
#define mem(a, b) memset(a, b, sizeof a)
#define sq(x) (x) * (x)
#define out return cout << "NE" << endl, 0
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops")
#pragma GCC optimize("no-stack-protector,fast-math,-ffp-contract=off")
const int mx = 5e4 + 10;
vector < int > v;
map < string , string > mp;
string a[mx], b[mx];
bool is(string s){
	return (s[0] >= '1' && s[0] <= '9');
}
int32_t main(){
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
	int n;
	cin >> n;
	for(int i = 0; i < n; i++)
		cin >> a[i];
	for(int i = 0; i < n; i++)
		cin >> b[i];
	for(int i = 0; i < n; i++){
		if(is(a[i]) && is(b[i])){
			if(a[i] != b[i])
				out;
		}
		else if(is(a[i])){
			if(mp.count(b[i]) && mp[b[i]] != a[i])
					out;
			mp[b[i]] = a[i];
		}
		else if(is(b[i])){
			if(mp.count(a[i]) && mp[a[i]] != b[i])
					out;
			mp[a[i]] = b[i];
		}
		else
			v.pb(i);
	}
	for(int u : v){
			int i = u;
			if(mp.count(a[i]) && mp.count(b[i])){
				if(mp[a[i]] != mp[b[i]])
					out;
			}
			else if(mp.count(a[i]))
				mp[b[i]] = mp[a[i]];
			else if(mp.count(b[i]))
				mp[a[i]] = mp[b[i]];
			else
				mp[a[i]] = mp[b[i]] = '1';
	}
	cout << "DA" << endl;
	return 0;
}

• Subtask 1: 14 / 14
• Subtask 2: 14 / 14
• Subtask 3: 14 / 14
• Subtask 4: 0 / 14
• Subtask 5: 0 / 14