Submission #1101470

#	Time	Username	Problem	Language	Result	Execution time	Memory
1101470		Anphat	K blocks (IZhO14_blocks)	C++14	100 / 100	339 ms	49916 KiB

blocks

#include <bits/stdc++.h>
using namespace std;

#define ll long long
#define f first
#define s second
#define sz(a) (int)(a.size())
#define all(a) a.begin(), a.end()
#define uni(a) sort(all(a)), a.resize(unique(all(a)) - a.begin())

typedef pair <int, int> ii;

const int N = 1e5 + 5;

int n, k, a[N], ans, l[N], r[N], dp[101][N];
stack <int> st;

int mn[N][20];
int *save;

void build() {
    for (int i = 0; i <= n; i++) mn[i][0] = save[i];
    for (int j = 1; (1 << j) <= n; 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]);
        }
    }
}

int get(int l, int r) {
    if (l > r) return 1e9;
    int lg = __lg(r - l + 1);
    return min(mn[l][lg], mn[r - (1 << lg) + 1][lg]);
}

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

    //freopen("a.inp", "r", stdin);
    //freopen("a.out", "w", stdout);

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

    a[0] = a[n + 1] = 1e9;
    st.push(0);

    for (int i = 1; i <= n; i++) {
        while (a[st.top()] <= a[i]) st.pop();
        l[i] = st.top();
        st.push(i);
    }

    while (sz(st)) st.pop();
    st.push(n + 1);

    for (int i = n; i >= 1; i--) {
        while (a[st.top()] <= a[i]) st.pop();
        r[i] = st.top();
        st.push(i);
    }

    memset(dp, 0x3f, sizeof(dp));
    dp[0][0] = 0; save = dp[0];

    for (int j = 1; j <= k; j++) {
        build();
        for (int i = 1; i <= n; i++) {
            dp[j][i] = get(l[i], i - 1) + a[i];
            dp[j][i] = min(dp[j][i], dp[j][l[i]]);
        }
        save = dp[j];
    }

    cout << dp[k][n] << '\n';

    return 0;
}

Compilation message (stderr)

blocks.cpp: In function 'void build()':
blocks.cpp:25:57: warning: suggest parentheses around '-' inside '<<' [-Wparentheses]
   25 |             mn[i][j] = min(mn[i][j - 1], mn[i + (1 << j - 1)][j - 1]);
      |                                                       ~~^~~

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