Submission #50032

#	Time	Username	Problem	Language	Result	Execution time	Memory
50032		Benq	Potemkin cycle (CEOI15_indcyc)	C++14	50 / 100	243 ms	2920 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> nex, ADJ[1001];
vi adj[1001], tmp[1001];
int comp[1001];

void finish(int x, int a1, int a2) {
    exit(0);
    FOR(i,1,N+1) dist[i] = MOD;
    queue<int> q; q.push(a1); dist[a1] = 0;
    
    while (sz(q)) {
        int t = q.front(); q.pop();
        if (t == x || (nex[t] && t != a1 && t != a2)) continue;
        
        for (int j: adj[t]) if (j <= x) {
            if (dist[j] != MOD) continue;
            dist[j] = dist[t] + 1;
            pre[j] = t;
            q.push(j);
        }
    }
    
    vi z;
    while (a2 != a1) {
        z.pb(a2);
        a2 = pre[a2];
    }
    z.pb(a1);
    z.pb(x);
    
    for (int i: z) cout << i << " ";
}

void dfs(int x, int lab, int mx) {
    if (comp[x]) return;
    comp[x] = lab;
    for (int i: adj[x]) 
        if (i < mx && !(nex[x] && nex[i]))
            dfs(i,lab,mx);
}

void test(int x) {
    FOR(i,1,N+1) comp[i] = 0, nex[i] = 0, tmp[i].clear();
    FOR(i,1,x) if (ADJ[x][i]) nex[i] = 1;
    
    FOR(i,1,x) if (nex[i]) {
        if (!comp[i]) dfs(i,i,x);
        tmp[comp[i]].pb(i);
    }
    
    FOR(i,1,x) F0R(j,sz(tmp[i])) FOR(k,j+1,sz(tmp[i])) 
        if (!ADJ[tmp[i][j]][tmp[i][k]])
            finish(x,tmp[i][j],tmp[i][k]);
}

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].pb(y), adj[y].pb(x);
        ADJ[x][y] = ADJ[y][x] = 1;
    }
    FORd(i,1,N+1) test(i);
    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: 0 / 10
• Subtask 6: 0 / 10
• Subtask 7: 0 / 10
• Subtask 8: 10 / 10
• Subtask 9: 0 / 10
• Subtask 10: 0 / 10