제출 #1098082

#제출 시각아이디문제언어결과실행 시간메모리
1098082anhthiK개의 묶음 (IZhO14_blocks)C++14
100 / 100
352 ms250120 KiB
#include <bits/stdc++.h>
 
using namespace std;
 
#define fi first
#define se second
#define ll long long
#define mp(x, y) make_pair(x, y)
#define sz(v) ((int) (v).size())
#define all(v) (v).begin(), (v).end()
#define MASK(i) (1LL << (i))
#define BIT(x, y) (((x) >> (y)) & 1)
#define pb push_back
#define heap priority_queue
#define max_rng *max_element
#define min_rng *min_element
#define rep(i, n) for(int i = 1, _n = (n); i <= _n; ++i)
#define forn(i, a, b) for(int i = (a), _b = (b); i <= _b; ++i)
#define ford(i, a, b) for(int i = (a), _b = (b); i >= _b; --i)

 
template <class X, class Y>
    inline bool maximize(X &x, Y y) {
        return (x < y ? x = y, true : false);
    }
 
template <class X, class Y>
    inline bool minimize(X &x, Y y) {
        return (x > y ? x = y, true : false);
    }
 
template <class X>
    inline void compress(vector<X> &a) {
        sort(all(a));
        a.resize(unique(all(a)) - a.begin());
    }
 
int ctz(ll x) { return __builtin_ctzll(x); }
int lg(ll x) { return 63 - __builtin_clzll(x); }
int popcount(ll x) { return __builtin_popcount(x); }
 
mt19937_64 rng(chrono::high_resolution_clock::now().time_since_epoch().count());
ll rand(ll l, ll r) {
    return l + abs((ll) rng()) % (r - l + 1);
}
 
const ll oo = (ll) 1e18 + 5;
const int inf = (ll) 1e9 + 7, mod = (ll) 1e9 + 7;
 
const int N = 3e5 + 5, LOG = 30;
 
void add(int &x, int y) { x += y; if (x >= mod) x -= mod; }
void sub(int &x, int y) { x -= y; if (x < 0) x += mod; }

int n, k;
ll a[N];

ll f[N][105];

int main(void) {
 
    ios_base::sync_with_stdio(0);
    cin.tie(0); cout.tie(0);
        
    cin >> n >> k;
    rep(i, n) cin >> a[i];

    memset(f, 0x3f, sizeof (f));    

    f[0][1] = 0;
    forn(i, 1, n) f[i][1] = max(f[i-1][1], a[i]);
    forn(j, 2, k) {
        vector<pair<ll, int>> st;
        forn(i, j, n) {
            ll res = f[i-1][j-1];
            while (sz(st) && a[st.back().second] <= a[i]) {
                minimize(res, st.back().first);
                st.pop_back();
            }
            f[i][j] = min(f[sz(st) ? st.back().second : 0][j], res + a[i]);
            st.pb({res, i});
        } 
    }

    cout << f[n][k];

    return 0;
}
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