Submission #86632

#	Time	Username	Problem	Language	Result	Execution time	Memory
86632		dimash241	Potemkin cycle (CEOI15_indcyc)	C++17	30 / 100	696 ms	37560 KiB

indcyc

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


#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 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)
// ROAD to...                                                                                                                                                                                                                Red

using namespace std;

inline bool isvowel (char c) {
	c = tolower(c);
    if (c == 'a' || c == 'e' || c == 'i' || c == 'y' || c == 'o' || c == 'u') return 1;
    return 0;
}

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 = 6e5 + 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;
vector < int > g[N];
vector < int > gr[N];
int has[5050][5050];
int x[N], y[N];
int p[N], pos[N];
int dp[N][5];

int main () {               
    cin >> n >> m ;
    For (i, 1, m) {
    	cin >> x[i] >> y[i];
    	g[x[i]].pb(y[i]);
    	g[y[i]].pb(x[i]);
    	has[x[i]][y[i]] = has[y[i]][x[i]] = 1;
    }

    for (int it = 1; it <= 100; ++it) {
    	for (int i = 1; i <= n; ++i) random_shuffle(every(g[i]));

    	For (i, 1, m) {
    		for (int j = 0; j < min(sz(g[x[i]]), 5); ++ j) {
    			if (y[i] == g[x[i]][j] || has[g[x[i]][j]][y[i]]) continue;
    			for (int l = 0; l < min(sz(g[y[i]]), 5); ++l) {
    				if (x[i] == g[y[i]][l]) continue;
    				if (!has[x[i]][g[y[i]][l]] && !has[g[x[i]][j]][y[i]] && has[g[x[i]][j]][g[y[i]][l]]) {
    					cout << g[x[i]][j] <<  ' ' << x[i] << ' ' << y[i] << ' ' << g[y[i]][l];
    					exit(0);
    				}
    			}
    		}
    	}
    }

    cout << "no";
    return Accepted;
}

// Coded By OB

• Subtask 1: 10 / 10
• Subtask 2: 10 / 10
• Subtask 3: 10 / 10
• Subtask 4: 0 / 10
• Subtask 5: 0 / 10
• Subtask 6: 0 / 10
• Subtask 7: 0 / 10
• Subtask 8: 0 / 10
• Subtask 9: 0 / 10
• Subtask 10: 0 / 10