Submission #50021

#	Time	Username	Problem	Language	Result	Execution time	Memory
50021		Benq	Potemkin cycle (CEOI15_indcyc)	C++14	70 / 100	1092 ms	3620 KiB

indcyc

#include <bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>

using namespace std;
using namespace __gnu_pbds;
 
typedef long long ll;
typedef long double ld;
typedef complex<ld> cd;

typedef pair<int, int> pi;
typedef pair<ll,ll> pl;
typedef pair<ld,ld> pd;

typedef vector<int> vi;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<pi> vpi;
typedef vector<pl> vpl;
typedef vector<cd> vcd;

template <class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>;

#define FOR(i, a, b) for (int i=a; i<(b); i++)
#define F0R(i, a) for (int i=0; i<(a); i++)
#define FORd(i,a,b) for (int i = (b)-1; i >= a; i--)
#define F0Rd(i,a) for (int i = (a)-1; i >= 0; i--)

#define sz(x) (int)(x).size()
#define mp make_pair
#define pb push_back
#define f first
#define s second
#define lb lower_bound
#define ub upper_bound
#define all(x) x.begin(), x.end()

const int MOD = 1000000007;
const ll INF = 1e18;
const int MX = 100001;

int N,R,dist[1001],pre[1001];
bitset<1001> adj[1001], ok;

void getPath(int x, int y) {
    queue<int> q; q.push(x); dist[x] = 0;
    while (sz(q)) {
        int t = q.front(); q.pop();
        //cout << "OH " << t << " " << adj[2][5] << "\n";
        FOR(i,1,N+1) if (ok[i] && adj[t][i]) {
            if (dist[i] != MOD || (t == x && i == y)) continue;
            //cout << "TRI " << t << " " << i << "\n";
            dist[i] = dist[t] + 1;
            pre[i] = t;
            q.push(i);
        }
    }
}
void test(int x, int y) {
    FOR(i,1,N+1) {
        ok[i] = 1;
        dist[i] = MOD;
    }
    FOR(i,1,N+1) if (adj[i][x] && adj[i][y]) ok[i] = 0;
    getPath(x,y);
    if (dist[y] == MOD) return;
    while (y != x) {
        cout << y << " ";
        y = pre[y];
    }
    cout << x;
    exit(0);
}

int main() {
    ios_base::sync_with_stdio(0); cin.tie(0);
    cin >> N >> R;
    F0R(i,R) {
        int x,y; cin >> x >> y;
        adj[x][y] = adj[y][x] = 1;
    }
    FOR(i,1,N+1) FOR(j,i+1,N+1) if (adj[i][j]) {
        test(i,j);
    }
    cout << "no";
}

// read the question correctly (is y a vowel? what are the exact constraints?)
// look out for SPECIAL CASES (n=1?) and overflow (ll vs int?)

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