Submission #750582

#TimeUsernameProblemLanguageResultExecution timeMemory
750582GrindMachineThe short shank; Redemption (BOI21_prison)C++17
15 / 100
2063 ms16952 KiB
// 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;

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){
        fill(all(dp2),inf2);

        rep1(i,n){
            // range add
            rep(j,lx[i]){
                dp1[j]++;
            }

            // range min
            rep(j,i){
                amin(dp2[i], dp1[j]);
            }
        }

        dp1 = dp2;
    }

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

int main()
{
    fastio;

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

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

    return 0;
}
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