제출 #750584

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
750584		GrindMachine	The short shank; Redemption (BOI21_prison)	C++17	35 / 100	2061 ms	12032 KiB

prison

// Om Namah Shivaya

#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) 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;

template<typename T>
struct lazysegtree {
    /*=======================================================*/

    struct data {
        ll a;
    };

    struct lazy {
        ll a;
    };

    data d_neutral = {inf2};
    lazy l_neutral = {0};

    void merge(data &curr, data &left, data &right) {
        curr.a = min(left.a, right.a);
    }

    void create(int x, int lx, int rx, T v) {
        tr[x].a = v;
    }

    void modify(int x, int lx, int rx, T v) {
        lz[x].a = v;
    }

    void propagate(int x, int lx, int rx) {
        ll v = lz[x].a;
        if(!v) return;

        tr[x].a += v;

        if(rx - lx > 1){
            lz[2*x+1].a += v;
            lz[2*x+2].a += v;
        }

        lz[x] = l_neutral;
    }

    /*=======================================================*/

    int siz = 1;
    vector<data> tr;
    vector<lazy> lz;

    lazysegtree() {

    }

    lazysegtree(int n) {
        while (siz < n) siz *= 2;
        tr.assign(2 * siz, d_neutral);
        lz.assign(2 * siz, l_neutral);
    }

    void build(vector<T> &a, int n, int x, int lx, int rx) {
        if (rx - lx == 1) {
            if (lx < n) {
                create(x, lx, rx, a[lx]);
            }

            return;
        }

        int mid = (lx + rx) / 2;

        build(a, n, 2 * x + 1, lx, mid);
        build(a, n, 2 * x + 2, mid, rx);

        merge(tr[x], tr[2 * x + 1], tr[2 * x + 2]);
    }

    void build(vector<T> &a, int n) {
        build(a, n, 0, 0, siz);
    }

    void rupd(int l, int r, T v, int x, int lx, int rx) {
        propagate(x, lx, rx);

        if (lx >= r or rx <= l) return;
        if (lx >= l and rx <= r) {
            modify(x, lx, rx, v);
            propagate(x, lx, rx);
            return;
        }

        int mid = (lx + rx) / 2;

        rupd(l, r, v, 2 * x + 1, lx, mid);
        rupd(l, r, v, 2 * x + 2, mid, rx);

        merge(tr[x], tr[2 * x + 1], tr[2 * x + 2]);
    }

    void rupd(int l, int r, T v) {
        rupd(l, r + 1, v, 0, 0, siz);
    }

    data query(int l, int r, int x, int lx, int rx) {
        propagate(x, lx, rx);

        if (lx >= r or rx <= l) return d_neutral;
        if (lx >= l and rx <= r) return tr[x];

        int mid = (lx + rx) / 2;

        data curr;
        data left = query(l, r, 2 * x + 1, lx, mid);
        data right = query(l, r, 2 * x + 2, mid, rx);

        merge(curr, left, right);
        return curr;
    }

    data query(int l, int r) {
        return query(l, r + 1, 0, 0, siz);
    }
};

void solve(int test_case)
{
    ll n,d,t; cin >> n >> d >> t;
    d++; // we need d+1 segs
    vector<ll> a(n+5);
    rep1(i,n) cin >> a[i];

    vector<ll> lx(n+5); // lx[i] = first pos to the left of i s.t if we start making ops from this pos, a[i] <= t
    rep1(i,n){
        auto b = a;
        for(int j = i; j <= n; ++j){
            if(j > i){
                amin(b[j], b[j-1] + 1);
            }

            if(b[j] <= t){
                amax(lx[j], (ll) i);
            }
        }
    }

    vector<ll> dp1(n+5,inf2), dp2(n+5,inf2);
    dp1[0] = 0;

    rep1(iter,d){
        lazysegtree<ll> st(n+5);        
        st.build(dp1,n+1);

        rep1(i,n){
            // range add
            st.rupd(0,lx[i]-1,1);

            // range min
            dp1[i] = st.query(0,i-1).a;
        }
    }

    ll ans = dp1[n];
    cout << ans << endl;
}

int main()
{
    fastio;

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

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

    return 0;
}

• Subtask 1: 15 / 15
• Subtask 2: 0 / 10
• Subtask 3: 20 / 20
• Subtask 4: 0 / 10
• Subtask 5: 0 / 25
• Subtask 6: 0 / 20