Submission #951785

#	Time	Username	Problem	Language	Result	Execution time	Memory
951785		EJIC_B_KEDAX	Copy and Paste 3 (JOI22_copypaste3)	C++17	57 / 100	1278 ms	596820 KiB

copypaste3

#ifdef LOCAL
    #define _GLIBCXX_DEBUG
#endif
#include <bits/stdc++.h>

#ifndef LOCAL
    // #pragma GCC optimize("O3")
    // #pragma GCC optimize("Ofast")
    // #pragma GCC optimize("unroll-loops")
    // #pragma GCC target("avx,avx2,bmi,bmi2,popcnt,lzcnt")
#endif
using namespace std;
using ll = long long;
using ld = long double;
#define x first
#define y second
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()

mt19937_64 mt(time(0));

void solve();
void init();

int32_t main() {
#ifndef LOCAL
    cin.tie(nullptr)->sync_with_stdio(false);
#endif
    cout << fixed << setprecision(30);
    init();
    int t = 1;
    // cin >> t;
    while (t--) {
        solve();
    }
}

char mem[1 << 28];
size_t curmem = 0;
void* operator new(size_t n) {
    curmem += n;
    return mem + curmem - n;
}
void operator delete(void*) {}
void operator delete(void*, size_t) {}

const int N = 2525, mod = 1000000321, p = 31;
ll dp[N][N];
int to[N][N], nw[N][N], hsh[N], pdeg[N], pinv[N];
vector<pair<int, int>> upd[N];

int add(int a, int b) {
    return a + b >= mod ? a + b - mod : a + b;
}

int sub(int a, int b) {
    return a >= b ? a - b : a - b + mod;
}

int mul(int a, int b) {
    return 1ll * a * b % mod;
}

int bin_pow(int a, int x) {
    int res = 1;
    while (x) {
        if (x & 1) {
            res = mul(res, a);
        }
        a = mul(a, a);
        x >>= 1;
    }
    return res;
}

int inv(int a) {
    return bin_pow(a, mod - 2);
}

void init() {
    pdeg[0] = 1;
    pdeg[1] = p;
    pinv[0] = 1;
    pinv[1] = inv(p);
    for (int i = 2; i < N; i++) {
        pdeg[i] = mul(pdeg[i - 1], p);
        pinv[i] = mul(pinv[i - 1], pinv[1]);
    }
}

int get_hash(int l, int r) {
    return mul(sub(hsh[r + 1], hsh[l]), pinv[l]);
}

struct segment_tree {
    vector<ll> st;
    int size;
    segment_tree(int sz = N) {
        st.resize(2 * sz, 0);
        size = sz;
    }
    void add(int i, ll v) {
        i += size;
        st[i] += v;
        i >>= 1;
        while (i) {
            st[i] = min(st[2 * i], st[2 * i + 1]);
            i >>= 1;
        }
    }
    ll get_min(int l, int r) {
        l += size;
        r += size;
        ll res = INT64_MAX;
        while (l <= r) {
            if (l & 1) {
                res = min(res, st[l++]);
            }
            if (~r & 1) {
                res = min(res, st[r--]);
            }
            l >>= 1;
            r >>= 1;
        }
        return res;
    }
};

segment_tree st[N];

void solve() {
    int n;
    cin >> n;
    string s;
    cin >> s;
    int a, b, c;
    cin >> a >> b >> c;
    hsh[0] = 0;
    for (int i = 0; i < n; i++) {
        hsh[i + 1] = add(hsh[i], mul(s[i] - 'a' + 1, pdeg[i]));
    }
    for (int l = 0; l < n; l++) {
        unordered_map<int, int> mp;
        for (int i = 0; i < n - l; i++) {
            int j = i + l;
            int nw = get_hash(i, j);
            if (mp.find(nw) == mp.end()) {
                to[i][j] = -1;
            } else {
                to[i][j] = mp[nw];
            }
            if (i >= l) {
                mp[get_hash(i - l, i)] = i - l;
            }
        }
    }
    for (int i = 0; i < n; i++) {
        dp[i][i] = a;
        // save[i][i] = b + c;
        st[i].add(i, b + c);
        nw[i][i] = to[i][i];
        if (to[i][i] != -1) {
            upd[i - to[i][i]].emplace_back(i, i);
        }
    }
    for (int l = 1; l < n; l++) {
        for (auto [i, j] : upd[l]) {
            int nxt = to[nw[i][j]][nw[i][j] + j - i];
            nw[i][j] = nxt;
            if (nxt >= 0) {
                upd[j - nxt].emplace_back(i, j);
            }
            // save[j][i] += b - 1ll * (j - i + 1) * a;
            st[j].add(i, c - 1ll * (j - i + 1) * a);
        }
        for (int i = 0; i < n - l; i++) {
            int j = i + l;
            dp[i][j] = min(1ll * (j - i + 1) * a + min(0ll, st[j].get_min(i + 1, j)), dp[i][j - 1] + a);
            st[j].add(i, dp[i][j] + b + c - 1ll * (j - i + 1) * a);
            nw[i][j] = to[i][j];
            if (nw[i][j] != -1) {
                upd[j - nw[i][j]].emplace_back(i, j);
            }
        }
    }
    // cout << to[3][3] << '\n';
    // for (int i = 0; i < n; i++) {
    //     for (int j = i; j < n; j++) {
    //         cout << dp[i][j] << ' ';
    //     }
    //     cout << '\n';
    // }
    cout << dp[0][n - 1] << '\n';
}

• Subtask 1: 1 / 1
• Subtask 2: 0 / 5
• Subtask 3: 14 / 14
• Subtask 4: 10 / 10
• Subtask 5: 32 / 32
• Subtask 6: 0 / 38