Submission #59349

#	Time	Username	Problem	Language	Result	Execution time	Memory
59349		Benq	Savez (COCI15_savez)	C++14	120 / 120	96 ms	20804 KiB

savez

#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 = 2000001;

vector<map<pair<char,char>,int>> tri;
int bes[MX];
int nex = 1, ans;

void ins(string t) {
    int cur = 0, res = 0;
    F0R(i,sz(t)) {
        pair<char,char> p = {t[i],t[sz(t)-1-i]};
        if (!tri[cur].count(p)) {
            tri[cur][p] = nex ++;
            tri.pb(map<pair<char,char>,int>());
        }
        cur = tri[cur][p];
        res = max(res,bes[cur]);
    }
    ans = max(ans,bes[cur] = res+1);
}

int N;

int main() {
    ios_base::sync_with_stdio(0); cin.tie(0);
    tri.pb(map<pair<char,char>,int>());
    cin >> N;
    F0R(i,N) {
        string s; cin >> s;
        ins(s);
    }
    cout << ans;
}

/* Look for:
* the exact constraints (multiple sets are too slow for n=10^6 :( ) 
* special cases (n=1?)
* overflow (ll vs int?)
* array bounds
*/

• Subtask 1: 12 / 12
• Subtask 2: 12 / 12
• Subtask 3: 12 / 12
• Subtask 4: 12 / 12
• Subtask 5: 12 / 12
• Subtask 6: 12 / 12
• Subtask 7: 12 / 12
• Subtask 8: 12 / 12
• Subtask 9: 12 / 12
• Subtask 10: 12 / 12