제출 #319014

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
319014		caoash	K개의 묶음 (IZhO14_blocks)	C++17	53 / 100	1089 ms	42340 KiB

blocks

#pragma GCC optimize("Ofast,unroll-loops")
#pragma GCC target("avx,avx2,sse,sse2")

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

using ll = long long;

using vi = vector<int>;
using vl = vector<ll>;
#define pb push_back
#define rsz resize
#define all(x) begin(x), end(x)
#define sz(x) (int)(x).size()

using pi = pair<int,int>;
#define f first
#define s second
#define mp make_pair

const int MX = 100000;
const int MOD = (int) (1e9 + 7);
const int INF = 987654321;

int dp[MX][101];

struct Node {
    int val;
};

template<int SZ> struct Seg {
    Node tree[4*SZ];

    Node merge(Node a, Node b){
        if (a.val < b.val) return a;
        return b;
    }

    void build(int v, int l, int r, int d) {
        if (l == r) {
            tree[v].val = dp[l][d];
        } else {
            int m = (l + r) / 2;
            build(2 * v + 1, l, m, d);
            build(2 * v + 2, m + 1, r, d);
            tree[v] = merge(tree[2 * v + 1], tree[2 * v + 2]);
        }
    }
    
    Node query(int v, int l, int r, int ql, int qr) {
        if (l > qr || r < ql) {
            Node init;
            init.val = INF;
            return init;
        } else if (l >= ql && r <= qr) {
            return tree[v];
        } else {
            int m = (l + r) / 2;
            Node a = query(2 * v + 1, l, m, ql, qr);
            Node b = query(2 * v + 2, m + 1, r, ql, qr);
            return merge(a, b);
        }
    }
};

Seg<MX> orz;

int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n, k; cin >> n >> k;
    vi a(n + 1), lft(n + 1);
    for (int i = 1; i <= n; i++) cin >> a[i];
    stack<pi> stk;
    for (int i = 1; i <= n; i++) {
        while (!stk.empty() && stk.top().f <= a[i]) {
            stk.pop();
        }
        lft[i] = (stk.empty() ? 0 : stk.top().s);
        stk.push(mp(a[i], i));
    }
    for (int i = 0; i <= n; i++) {
        for (int j = 0; j <= k; j++) {
            dp[i][j] = INF;
        }
    }
    dp[0][0] = 0;
    for (int j = 0; j <= k; j++) {
        for (int i = 1; i <= n; i++) {
            if (lft[i] != 0) dp[i][j] = min(dp[i][j], dp[lft[i]][j]);
            if (j) dp[i][j] = min(dp[i][j], orz.query(0, 0, n, lft[i], i - 1).val + a[i]);
        }
        orz.build(0, 0, n, j);
    }
    cout << dp[n][k] << '\n';
}

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