Submission #978796

#	Time	Username	Problem	Language	Result	Execution time	Memory
978796		LOLOLO	K blocks (IZhO14_blocks)	C++17	53 / 100	1065 ms	97996 KiB

blocks

#include <bits/stdc++.h>
#define ll long long
using namespace std;
 
#define           f     first
#define           s     second
#define           pb    push_back
#define           ep    emplace
#define           eb    emplace_back
#define           lb    lower_bound
#define           ub    upper_bound
#define       all(x)    x.begin(), x.end()
#define      rall(x)    x.rbegin(), x.rend()
#define   uniquev(v)    sort(all(v)), (v).resize(unique(all(v)) - (v).begin())
#define     mem(f,x)    memset(f , x , sizeof(f))
#define        sz(x)    (int)(x).size()
#define  __lcm(a, b)    (1ll * ((a) / __gcd((a), (b))) * (b))
#define          mxx    *max_element
#define          mnn    *min_element
#define    cntbit(x)    __builtin_popcountll(x)
#define       len(x)    (int)(x.length())
 
const int N = 1e5 + 10;
ll f[N][101], a[N];
ll sp[N][20], n, k;
int lg[N];

void build(int id) {
    for (int i = 0; i <= n; i++) {
        sp[i][0] = f[i][id];
    }

    for (int j = 1; j < 20; j++) {
        for (int i = 0; i + (1 << j) <= n + 1; i++) {
            sp[i][j] = min(sp[i][j - 1], sp[i + (1 << (j - 1))][j - 1]);
        }
    }
}

ll range(int l, int r, int id) {
    if (l > r)
        return 1e16;

    int t = lg[r - l + 1];
    return min(sp[l][t], sp[r - (1 << t) + 1][t]);
}

int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);

    for (int i = 2; i < N; i++)
        lg[i] = lg[i / 2] + 1;

    mem(f, 0x3f);
    cin >> n >> k;

    f[0][0] = 0;
    for (int i = 1; i <= n; i++) {
        cin >> a[i];
    }

    for (int j = 1; j <= k; j++) {
        build(j - 1);
        stack <int> st;
        vector <ll> pr;

        for (int i = 0; i <= n; i++) {
            while (sz(st) && a[i] >= a[st.top()]) {
                st.pop();
                pr.pop_back();
            }

            int lst = sz(st) ? st.top() : 0;
            ll nxt = range(lst, i - 1, j - 1) + a[i];
            if (sz(pr))
                nxt = min(nxt, pr.back());

            f[i][j] = nxt;
            st.push(i);
            ll v = sz(pr) ? pr.back() : 1e18;
            pr.pb(min(v, range(lst, i - 1, j - 1) + a[i])); 
        }
    }

    cout << f[n][k] << '\n';
} 

// 2 2
// 1 2

• Subtask 1: 14 / 14
• Subtask 2: 18 / 18
• Subtask 3: 21 / 21
• Subtask 4: 0 / 47