Submission #681759

#	Time	Username	Problem	Language	Result	Execution time	Memory
681759		pragmatist	K blocks (IZhO14_blocks)	C++17	0 / 100	4 ms	4248 KiB

blocks

/*
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("O3")
#pragma GCC target ("avx2")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("unroll-loops")
*/

#include<bits/stdc++.h>
 
#define ld long double
#define sz(v) (int)v.size()
#define ll long long
#define pb push_back
#define x first
#define y second
#define all(v) v.begin(), v.end()
#define rall(v) v.rbegin(), v.rend()
#define nl "\n"

using namespace std;
using pii = pair<int, int>;

const int N = (int)1e6 + 7; // make sure this is right
const int M = (int)1e6 + 7;
const int inf = (int)4e9 + 7;
const ll INF = (ll)3e18 + 7; 
const ll MOD = (ll)998244353; // make sure this is right

bool bit(int x, int i) {
	return x >> i & 1;
}

int sum(int x, int y) {
	x += y;
	if(x >= MOD) x -= MOD;
	return x;
}

pii dir[] = {{1, 0}, {0, 1}, {-1, 0}, {0, -1}};

int n, k, a[N];
int lo[N], mx[N][20];
ll dp[N][2];

int get(int l, int r) {
	int x = lo[r - l + 1];
	return max(mx[l][x], mx[r - (1 << x) + 1][x]);
}

void calc(int l, int r, int optl, int optr) {
	if(l > r) return;
	int mid = (l + r) >> 1, optk = -1;
	for(int i = optl; i <= min(mid, optr); ++i) {
		ll cur = dp[i - 1][0] + get(i, mid);
		if(cur < dp[mid][1]) {
			dp[mid][1] = cur;
			optk = i;
		}
	}
	calc(l, mid - 1, optl, optk);
	calc(mid + 1, r, optk, optr);
}

void solve() {
	cin >> n >> k;
	for(int i = 1; i <= n; ++i) {
		cin >> a[i];
		mx[i][0] = a[i];
	}
	for(int i = 1; i < 20; ++i) {
		for(int j = 1; j <= n - (1 << i) + 1; ++j) {
			mx[j][i] = max(mx[j][i - 1], mx[j + (1 << (i - 1))][i - 1]);
		}
	}
	for(int i = 1; i <= n; ++i) dp[i][1] = dp[i][0] = INF;
	for(int i = 1; i <= k; ++i) {
		calc(1, n, 1, n);
	    for(int j = 1; j <= n; ++j) {
			dp[j][0] = dp[j][1];
			dp[j][1] = INF;
		}          
		dp[0][0] = INF;
	}
	cout << dp[n][0];
}

signed main() {	
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	for(int i = 2; i <= 1e6; ++i) lo[i] = lo[i / 2] + 1;
	int test = 1;
	//cin >> test;
	for(int i = 1; i <= test; ++i) {
	 	//cout << "Case #" << i << ": ";
		solve();
	}
	return 0;
}

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