Submission #555778

#TimeUsernameProblemLanguageResultExecution timeMemory
555778AriaHCopy and Paste 3 (JOI22_copypaste3)C++17
100 / 100
723 ms83188 KiB
/* Im the best and i work like that */

#pragma GCC optimize("Ofast")

#include <bits/stdc++.h>

using namespace std;

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

#define F first
#define S second
#define all(x) x.begin(),x.end()
#define Mp make_pair
#define point complex
#define endl '\n'
#define SZ(x) (int)x.size()
#define fast_io ios::sync_with_stdio(0), cin.tie(0), cout.tie(0)
#define file_io freopen("input.txt", "r+", stdin); freopen("output.txt", "w+", stdout);
#define mashtali return cout << "Hello, World!", 0;
#define int ll

const int N = 2510;
const int LOG = 20;
const ll inf = 8e18;
const double pi = acos(-1);
const ld eps = 1e-18;
const ld one = 1.;
const ll mod = 16227933338363;
const ll base = 1e3 + 9;

ll pw(ll a, ll b, ll M, ll ret = 1) { if(a == 0) return 0; a %= M; while(b) { ret = (b & 1? ret * a % M : ret), a = a * a % M, b >>= 1; } return ret % M; }

mt19937 rng(time(nullptr));

int n, par[N];

ll A, B, C, Hsh[N][N], P[N], dp[N][N]; /// dp -> how much we can save

string s;

inline ll get(int l, int r)
{
    return Hsh[l][r];
}

int32_t main()
{
	fast_io;
    cin >> n >> s >> A >> B >> C;
    s = "." + s;
    for(int i = 1; i <= n; i ++)
    {
        for(int j = i; j <= n; j ++)
        {
            Hsh[i][j] = (Hsh[i][j - 1] * base + s[j]) % mod;
        }
    }
    for(int len = 1; len <= n; len ++)
    {
        memset(par, -1, sizeof par);
        vector < pll > vec;
        for(int i = len; i <= n; i ++)
        {
            vec.push_back(Mp(get(i - len + 1, i), i));
        }
        for(int i = 1; i <= n; i ++)
        {
            dp[i][i + len - 1] = max({dp[i][i + len - 1], dp[i + 1][i + len - 1], dp[i][i + len - 2]});
        }
        sort(all(vec));
        for(int i = len; i <= n; i ++)
        {
            ll cu = get(i - len + 1, i);
            int id = lower_bound(all(vec), Mp(cu, 1ll * i + len)) - vec.begin();
            if(id < SZ(vec) && vec[id].F == cu)
            {
                par[i] = vec[id].S;
            }
        }
        for(int i = len; i <= n; i ++)
        {
            int nxt = par[i], cnt = 1;
            ll cu = dp[i - len + 1][i];
            while(nxt != -1)
            {
                cnt ++;
                dp[i - len + 1][nxt] = max(dp[i - len + 1][nxt], cu + 1ll * len * (cnt - 1) * A - B - 1ll * cnt * C);
                nxt = par[nxt];
            }
        }
    }
    cout << 1ll * n * A - dp[1][n];
	return 0;
}

/* check corner case(n = 1?), watch for negetive index or overflow */
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