제출 #1208114

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1208114		gatapdev	K개의 묶음 (IZhO14_blocks)	C++20	53 / 100	1095 ms	4444 KiB

blocks

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const ll INF = LLONG_MAX / 4;

// Segment tree supporting range-add and range-min queries
struct SegTree {
    int n;
    vector<ll> st, lazy;
    SegTree(int _n) : n(_n), st(4*n, INF), lazy(4*n, 0) {}
    void apply(int p, int l, int r, ll v) {
        st[p] += v;
        lazy[p] += v;
    }
    void push(int p, int l, int r) {
        if (lazy[p] != 0) {
            int m = (l + r) / 2;
            apply(p<<1, l, m, lazy[p]);
            apply(p<<1|1, m+1, r, lazy[p]);
            lazy[p] = 0;
        }
    }
    void pull(int p) {
        st[p] = min(st[p<<1], st[p<<1|1]);
    }
    void build(int p, int l, int r, const vector<ll>& base) {
        if (l == r) {
            st[p] = base[l];
            return;
        }
        int m = (l + r) / 2;
        build(p<<1, l, m, base);
        build(p<<1|1, m+1, r, base);
        pull(p);
    }
    void range_add(int p, int l, int r, int i, int j, ll v) {
        if (i > r || j < l) return;
        if (i <= l && r <= j) {
            apply(p, l, r, v);
            return;
        }
        push(p, l, r);
        int m = (l + r) / 2;
        range_add(p<<1, l, m, i, j, v);
        range_add(p<<1|1, m+1, r, i, j, v);
        pull(p);
    }
    ll range_min(int p, int l, int r, int i, int j) {
        if (i > r || j < l) return INF;
        if (i <= l && r <= j) return st[p];
        push(p, l, r);
        int m = (l + r) / 2;
        return min(range_min(p<<1, l, m, i, j), range_min(p<<1|1, m+1, r, i, j));
    }
};

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int n, K;
    cin >> n >> K;
    vector<ll> a(n+1);
    for (int i = 1; i <= n; i++) cin >> a[i];

    vector<ll> dp_prev(n+1, INF), dp_cur(n+1, INF);
    dp_prev[0] = 0;
    // base case k=1: prefix max
    for (int i = 1; i <= n; i++) dp_prev[i] = max(dp_prev[i-1], a[i]);

    for (int k = 2; k <= K; k++) {
        SegTree st(n+1);
        st.build(1, 0, n, dp_prev);

        vector<int> stk, L(n+1);
        stk.reserve(n);
        fill(dp_cur.begin(), dp_cur.end(), INF);

        for (int i = k; i <= n; i++) {
            // initial left boundary
            L[i] = i;
            // pop smaller elements
            while (!stk.empty() && a[stk.back()] <= a[i]) {
                int t = stk.back(); stk.pop_back();
                int lt = L[t];
                // remove old max contribution
                st.range_add(1, 0, n, lt-1, t-1, -a[t]);
                L[i] = min(L[i], lt);
            }
            // add new max contribution
            st.range_add(1, 0, n, L[i]-1, i-1, a[i]);
            stk.push_back(i);
            // compute dp
            dp_cur[i] = st.range_min(1, 0, n, k-1, i-1);
        }
        dp_prev.swap(dp_cur);
    }

    cout << dp_prev[n] << '\n';
    return 0;
}

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