Submission #39033

#	Time	Username	Problem	Language	Result	Execution time	Memory
39033		RockyB	K blocks (IZhO14_blocks)	C++14	100 / 100	516 ms	43424 KiB

blocks

/// In The Name Of God

#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,sse3,sse4,popcnt,abm,mmx")

#include <bits/stdc++.h>

#define f first
#define s second

#define pb push_back
#define pp pop_back
#define mp make_pair

#define sz(x) (int)x.size()
#define sqr(x) ((x) * 1ll * (x))
#define all(x) x.begin(), x.end()

#define Kazakhstan ios_base :: sync_with_stdio(0), cin.tie(0), cout.tie(0);

#define nl '\n'
#define ioi exit(0);

typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;

const int N = (int)1e5 + 7, inf = (int)1e9 + 7, mod = (int)1e9 + 7;
const ll linf = (ll)1e18 + 7;
const int dx[] = {-1, 0, 1, 0, 1, -1, -1, 1}, dy[] = {0, 1, 0, -1, 1, -1, 1, -1};

using namespace std;

int get_int() {
	char x = getchar();
	while (!isdigit(x)) x = getchar();
	int res = 0;
	while (isdigit(x)) res = res * 10 + x - '0', x = getchar();
	return res;
}

int n, k;
int a[N];
int dp[101][N];

/*int get_min(int o, int l, int r) {
	int res = dp[o][l];
	for (int i = l + 1; i <= r; i++) {
		res = min(res, dp[o][i]);
	}
	return res;
}*/

int t[N << 1];
void build(int o) {
	for (int i = 0; i <= n; i++) {
		t[i + n] = dp[o][i];
	}
	for (int i = n - 1; i >= 0; i--) {
		t[i] = min(t[i << 1], t[i << 1 | 1]);
	}
}
int get_min(int l, int r) {
	int res = inf;
	for (l += n, r += n + 1; l < r; l >>= 1, r >>= 1) {
		if (l & 1) res = min(res, t[l++]);
		if (r & 1) res = min(res, t[--r]);
	}
	return res;
}

int sz;
pair <int, int> c[N];
int main() {
	#ifdef IOI2018
		freopen ("in.txt", "r", stdin);
	#endif
	n = get_int(), k = get_int();
	for (int i = 1; i <= n; i++) 
		a[i] = get_int();
	memset(dp, 0x3f3f, sizeof(dp));
	dp[0][0] = 0;
	a[0] = inf;
	for (int o = 1; o <= k; o++) {
		sz = 1;
		c[sz] = {0, INT_MAX};
		build(o - 1);
		for (int i = 1; i <= n; i++) {
			while (sz && a[i] >= a[c[sz].f]) {
				sz--;
			}
			c[++sz].f = i;
			c[sz].s = get_min(c[sz - 1].f, c[sz].f - 1) + a[c[sz].f];
			c[sz].s = min(c[sz].s, c[sz - 1].s);
			dp[o][i] = c[sz].s;
		}
	}
	cout << dp[k][n];
	ioi
}

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