제출 #343660

#제출 시각아이디문제언어결과실행 시간메모리
343660_aniK개의 묶음 (IZhO14_blocks)C++17
53 / 100
1089 ms97644 KiB
#define BUGO(x) cerr << #x << " = " << (x) << '\n';
#include <iostream>
#include <algorithm>
#include <stack>
using namespace std;
using ll = long long;
const int N = 100'002;
const int K = 20;
const ll inf = 1'000'000'000'000'000'000LL;
ll dp[N][102], a[N];
int lg[N];
ll mn[N][K];
struct el {
    ll x; int ind;
};
bool operator<(const el& a, const el& b) {
    return a.x < b.x;
}
void Build(int n, int curk)
{
    int k = lg[n];
    for (int i = 0; i < n; i++)
        mn[i][0] = dp[i + 1][curk];
    for (int j = 1; j <= k; j++)
        for (int i = 0; i + (1 << j) - 1 <= n; i++)
            mn[i][j] = min(mn[i][j - 1], mn[i + (1 << (j - 1))][j - 1]);
}
ll Mn(int l, int r)
{
    l--; r--;
    int j = lg[r - l + 1];
    return min(mn[l][j], mn[r - (1 << j) + 1][j]);
}
int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    int n, k;
    cin >> n >> k;
    lg[1] = 0;
    for (int i = 2; i <= n; i++)
        lg[i] = lg[i / 2] + 1;
    for (int i = 1; i <= n; i++)
        cin >> a[i];
    for (int i = 0; i <= n; i++)
        for (int curk = 1; curk <= k; curk++)
            dp[i][curk] = inf;

    for (int curk = 1; curk <= k; curk++) {
        stack<el> x;
        x.push({ inf, 0 });
        for (int i = 1; i <= n; i++) {
            while (x.top().x < a[i])x.pop();
            el prev = x.top();
            x.push({ a[i], i });
            if (i < curk) {
                continue;
            }
            if (curk == 1)
                dp[i][curk] = ((i - 1 != 0) ? max(dp[i - 1][1], a[i]) : a[i]);
            else if (prev.ind == 0) {
                dp[i][curk] = ((i - 1 == 0) ? 0 : Mn(1, i - 1)) + a[i];
            }
            else dp[i][curk] = min(dp[prev.ind][curk], Mn(prev.ind, i - 1) + a[i]);
        }
        Build(n, curk);
    }

    /*for (int i = 0; i <= n; i++)
    {
        for (int curk = 0; curk <= k; curk++)
            cerr << dp[i][curk] << ' ';
        cerr << '\n';
    }*/
    cout << dp[n][k] << '\n';
    return 0;
}
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