제출 #1199511

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1199511		A_M_Namdar	K개의 묶음 (IZhO14_blocks)	C++20	0 / 100	36 ms	86344 KiB

blocks

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

const long long N = 1e5 + 10, M = 100 + 10, inf = 1e9 + 100;
long long n, k, a[N], dp[N][M], dp2[N];
struct node {
	long long mini, lazy;
} seg[N << 2];

void build(long long l, long long r, long long t, long long id) {
	seg[id].lazy = 0;
	if (r - l == 1) {
		seg[id].mini = dp[l][t];
		return;
	}
	long long mid = (l + r) / 2;
	build(l, mid, t, id << 1), build(mid, r, t, id << 1 | 1);
	seg[id].mini = min(seg[id << 1].mini, seg[id << 1 | 1].mini);
}

void shift(long long id) {
	seg[id << 1].mini += seg[id].lazy, seg[id << 1 | 1].mini += seg[id].lazy;
	seg[id << 1].lazy += seg[id].lazy, seg[id << 1 | 1].lazy += seg[id].lazy;
	seg[id].lazy = 0;
}

void add(long long l1, long long r1, long long l2, long long r2, long long v, long long id) {
	if (l1 >= r2 || r1 <= l2) return;
	if (l1 >= l2 && r1 <= r2) {
		seg[id].mini += v, seg[id].lazy += v;
		return;
	}
	shift(id);
	long long mid = (l1 + r1) / 2;
	add(l1, mid, l2, r2, v, id << 1), add(mid, r1, l2, r2, v, id << 1 | 1);
	seg[id].mini = min(seg[id << 1].mini, seg[id << 1 | 1].mini);
}

long long get(long long l1, long long r1, long long l2, long long r2, long long id) {
	if (l1 >= r2 || r1 <= l2) return inf;
	if (l1 >= l2 && r1 <= r2) return seg[id].mini;
	shift(id);
	long long mid = (l1 + r1) / 2;
	return min(get(l1, mid, l2, r2, id << 1), get(mid, r1, l2, r2, id << 1 | 1));
}

void input() {
	cin >> n >> k;
	for (long long i = 0; i < n; i++) 
		cin >> a[i];
}

void solve() {
	for (long long i = 0; i < n; i++) {
		dp2[i] = i - 1;
		while (dp2[i] > -1 && a[dp2[i]] <= a[i]) 
			dp2[i] = dp2[dp2[i]];
	}
	memset(dp, 63, sizeof dp);
	long long tmp = 0;
	for (long long i = 0; i < n; i++) {
		tmp = max(tmp, a[i]);
		dp[i][1] = tmp;
	}
	for (long long j = 2; j <= k; j++) {
		build(0, n, j - 1, 1);
		for (long long i = 0; i < n; i++) {
			add(0, n, i, i + 1, a[i], 1);
			long long p = i - 1;
			while (p > -1 && a[p] < a[i]) {
				add(0, n, dp2[p] + 1, p + 1, a[i] - a[p], 1);
				p = dp2[p];
			}
			dp[i][j] = get(0, n, 0, i, 1);
		}
	}
	cout << dp[n - 1][k];
}

int main() {
	ios:: sync_with_stdio(0), cin.tie(0), cout.tie(0);
	input();
	solve();
}

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