Submission #564073

#	Time	Username	Problem	Language	Result	Execution time	Memory
564073		perchuts	Savez (COCI15_savez)	C++17	0 / 120	1082 ms	52428 KiB

savez

#include <bits/stdc++.h>
#define all(x) x.begin(), x.end()
#define sz(x) (int) x.size()
#define endl '\n'
#define pb push_back
#define _ ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);

using namespace std;

using ll = long long;
using ull = unsigned long long;
using ii = pair<int,int>;
using iii = tuple<int,int,int>;

const int inf = 2e9+1;
const ll mod = 1e9+7;
const ll base = 101;
const int maxn = 2e6+100;

template<typename X, typename Y> bool ckmin(X& x, const Y& y) { return (y < x) ? (x=y,1):0; }
template<typename X, typename Y> bool ckmax(X& x, const Y& y) { return (x < y) ? (x=y,1):0; }

bool cmp(pair<int,string>a,pair<int,string>b){
    if(a.first==b.first)return 1;
    return a.first>b.first;
}

vector<vector<ll>>hsh, revHsh;
vector<pair<int,string>>v;
ll mul[maxn];

void computeHash(int idx){
    int n = v[idx].first;
    hsh[idx].pb(ll(v[idx].second[0]-'A'));
    for(int i=1;i<n;++i){
        hsh[idx].pb(hsh[idx].back()*base+ll(v[idx].second[i]-'A'));
        hsh[idx].back() %= mod;
    }
    revHsh[idx].pb(ll(v[idx].second[n-1]-'A'));
    for(int i=n-2;i>=0;--i){
        revHsh[idx].pb(revHsh[idx].back()+mul[n-1-i]*ll(v[idx].second[i]-'A'));
        revHsh[idx].back() %= mod;
    }
}

map<ll,int>dp;

int main(){_
    int n;cin>>n;
    v.resize(n);
    for(int i=0;i<n;++i){
        cin>>v[i].second;
        v[i].first = sz(v[i].second);
    }
    mul[0] = 1ll;
    for(int i=1;i<maxn;++i)mul[i] = (mul[i-1]*base)%mod;
    sort(all(v), cmp);
    hsh.resize(n), revHsh.resize(n);
    int best = 0;
    for(int i=0;i<n;++i){
        computeHash(i);
        ll me = hsh[i][v[i].first-1];
        int res = dp[me]+1;
        for(int j=0;j<v[i].first;++j){
            //termino aqui
            if(hsh[i][j]!=revHsh[i][j])continue;
            ckmax(dp[hsh[i][j]], res); 
        }
        ckmax(best, res);
    }

    cout<<best<<endl;
}

• Subtask 1: 0 / 12
• Subtask 2: 0 / 12
• Subtask 3: 0 / 12
• Subtask 4: 0 / 12
• Subtask 5: 0 / 12
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• Subtask 10: 0 / 12