Submission #241369

#	Time	Username	Problem	Language	Result	Execution time	Memory
241369		ne4eHbKa	COCI17_cezar (COCI17_cezar)	C++17	0 / 50	6 ms	512 KiB

cezar

//{ <defines>
#include <bits/stdc++.h>
using namespace std;

//#pragma comment(linker, "/stack:200000000")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
//#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math,-O3")

#define fr(i, n) for(int i = 0; i < n; ++i)
#define fo(n) fr(i, n)

#define re return
#define ef else if
#define ifn(x) if(!(x))
#define _  << ' ' <<

#define ft first
#define sd second
#define ve vector
#define pb push_back
#define eb emplace_back

#define sz(x) int((x).size())
#define bnd(x) x.begin(), x.end()
#define clr(x, y) memset((x), (y), sizeof (x))

typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef ve<int> vi;

inline ll time() {re chrono :: system_clock().now().time_since_epoch().count();}
mt19937 rnd(time());
mt19937_64 RND(time());

template<typename t> inline void umin(t &a, t b) {a = min(a, b);}
template<typename t> inline void umax(t &a, t b) {a = max(a, b);}

int md = 998244353;

inline int m_add(int&a, int b) {a += b; if(a < 0) a += md; if(a >= md) a -= md; re a;}
inline int m_sum(int a, int b) {a += b; if(a < 0) a += md; if(a >= md) a -= md; re a;}
inline int m_mul(int&a, int b) {re a = 1ll * a * b % md;}
inline int m_prod(int a, int b) {re 1ll * a * b % md;}

int m_bpow(ll A, ll b) {
    int a = A % md;
    ll ans = 1;
    for(ll p = 1ll << 63 - __builtin_clzll(b); p; p >>= 1) {
        (ans *= ans) %= md;
        if(p & b)
            (ans *= a) %= md;
    }
    re ans;
}

//const ld pi = arg(complex<ld>(-1, 0));
//const ld pi2 = pi + pi;
const int oo = 2e9;
const ll OO = 4e18;
//} </defines>
const int N = 2e5 + 5;
const int A = 26;

vi g[A];
map<int, int> h[A];

void solve() {
    int n; cin >> n;
    string s[n]; fo(n) cin >> s[i];
    int a[n]; for(int &i : a) cin >> i, --i;
    ve< pair<int, int> > e;
    fo(A) g[i].clear(), h[i].clear();
    for(int i = 1; i < n; ++i) {
        int l = a[i - 1], r = a[i];
        int j = 0;
        int na = sz(s[l]), nb = sz(s[r]);
        for(; j < min(na, nb) && s[l][j] == s[r][j]; ++j);
        if(j == min(na, nb)) {
            if(na > nb) return void(cout << "NE\n");
            continue;
        }
        l = s[l][j] - 'a', r = s[r][j] - 'a';
        g[l].pb(r);
        h[r][l]++;
    }
    queue<int> q;
    vector<bool> mk(A, false);
    for(int i = 0; i < A; ++i) if(h[i].empty()) q.push(i);
    vi z;
    for(; !q.empty(); q.pop()) {
        int v = q.front();
        if(mk[v]) continue;
        mk[v] = true;
        z.push_back(v);
        for(int u : g[v]) {
            ifn(--h[u][v]) h[u].erase(v);
            if(h[u].empty()) q.push(u);
        }
    }
    if(z.size() < A) return void(cout << "NE\n");
    ve<char> ans(A);
    cout << "DA\n";
    fo(A) ans[z[i]] = i + 'a';
    for(char i : ans) cout << i;
    cout << endl;
}

int main() {
#ifdef _LOCAL
    freopen("in.txt", "r", stdin);
    int tests; cin >> tests;
    for(int test = 1; test <= tests; ++test) {
		cerr << "case #" << test << endl;
        solve();
        cerr << endl;
    }
#else
//    freopen("input.txt", "r", stdin); freopen("output.txt", "w", stdout);
    ios_base :: sync_with_stdio(0); cin.tie(0); cout.tie(0);
    solve();
#endif
    return 0;
}

Compilation message (stderr)

cezar.cpp: In function 'int m_bpow(ll, ll)':
cezar.cpp:50:26: warning: suggest parentheses around '-' inside '<<' [-Wparentheses]
     for(ll p = 1ll << 63 - __builtin_clzll(b); p; p >>= 1) {
                       ~~~^~~~~~~~~~~~~~~~~~~~

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