Submission #950174

#TimeUsernameProblemLanguageResultExecution timeMemory
950174GrindMachineVim (BOI13_vim)C++17
1.39 / 100
880 ms524288 KiB
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace std;
using namespace __gnu_pbds;

template<typename T> using Tree = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
typedef long long int ll;
typedef long double ld;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;

#define fastio ios_base::sync_with_stdio(false); cin.tie(NULL)
#define pb push_back
#define endl '\n'
#define sz(a) (int)a.size()
#define setbits(x) __builtin_popcountll(x)
#define ff first
#define ss second
#define conts continue
#define ceil2(x,y) ((x+y-1)/(y))
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define yes cout << "Yes" << endl
#define no cout << "No" << endl

#define rep(i,n) for(int i = 0; i < n; ++i)
#define rep1(i,n) for(int i = 1; i <= n; ++i)
#define rev(i,s,e) for(int i = s; i >= e; --i)
#define trav(i,a) for(auto &i : a)

template<typename T>
void amin(T &a, T b) {
    a = min(a,b);
}

template<typename T>
void amax(T &a, T b) {
    a = max(a,b);
}

#ifdef LOCAL
#include "debug.h"
#else
#define debug(x) 42
#endif

/*



*/

const int MOD = 1e9 + 7;
const int N = 1e5 + 5;
const int inf1 = int(1e9) + 5;
const ll inf2 = ll(1e18) + 5;

void solve(int test_case)
{
    ll n; cin >> n;
    string s; cin >> s;
    s = "$" + s;

    ll nxt[n+5][10];
    memset(nxt,0x3f,sizeof nxt);
    array<ll,10> closest;
    closest.fill(inf2);

    rev(i,n,1){
        rep(j,10){
            nxt[i][j] = closest[j];
        }
        closest[s[i]-'a'] = i;
    }

    vector<ll> p(n+5);
    rep1(i,n) p[i] = p[i-1]+(s[i] == 'e');
    ll m = p[n];

    ll dp[n+5][m+5];
    memset(dp,0x3f,sizeof dp);
    dp[1][0] = 0;

    rep1(i,n){
        rep(j,m+1){
            rep(ch,10){
                if(ch == 4) conts;
                // if(dp[i][j]) conts;

                ll k = nxt[i][ch];
                if(k > n) conts;
                ll cnt = p[k]-p[i];

                for(int x = i; x <= min(nxt[i][4],k); ++x){
                    amin(dp[x][p[k]],dp[i][j]+1+cnt*2+k-x);
                }

                if(nxt[i][4]+1 <= k){
                    ll x = nxt[i][4]+1;
                    ll two_cnt = cnt-1;
                    ll one_cnt = k-(x-1)+1-two_cnt;
                    amin(dp[x][p[k]],dp[i][j]+1+cnt*two_cnt+one_cnt);
                }
            }
        }
    }

    ll ans = inf2;
    rep1(i,n){
        amin(ans,dp[i][m]);
    }

    ans += m-1;
    cout << ans << endl;
}

int main()
{
    fastio;

    int t = 1;
    // cin >> t;

    rep1(i, t) {
        solve(i);
    }

    return 0;
}
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