제출 #977548

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
977548		josanneo22	수열 (APIO14_sequence)	C++17	0 / 100	10 ms	7000 KiB

sequence

#pragma GCC optimize("O3")

#include <bits/stdc++.h>
using namespace std;
using i64 = long long;

#define L(i, j, k) for (int i = (j); i <= (k); ++i)
#define R(i, j, k) for (int i = (j); i >= (k); --i)
#define all(x) x.begin(), x.end()

const int nax = 100050;
const int kax = 204;

int N, K;
int a[nax];
int from[kax][nax];
i64 dp[2][nax];
i64 pre[nax];
int q[nax];
int L = 1, R = 1;

int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr); cout.tie(nullptr);

	cin >> N >> K;
	L(i, 1, N) {
		cin >> a[i];
		pre[i] = pre[i - 1] + a[i];
	}
	L(i, 0, N) dp[0][i] = 0;
	L(k, 1, K) {
		q[R++] = 0;
		int nw = k & 1;
		int pv = nw ^ 1;
		auto Y = [&](int x) { return dp[pv][x]; };
		auto X = [&](int x) { return pre[x]; };
		auto dY = [&](int x, int y) { return Y(x) - Y(y); };
		auto dX = [&](int x, int y) { return X(x) - X(y); };
		auto DP = [&](int x, int y) { return dp[pv][y] + (pre[x] - pre[y]) * (pre[N] - pre[x]); };
		L(i, 1, N) {
			while (R - L > 1 && dY(q[L + 1], q[L]) >= (pre[N] - pre[i]) * dX(q[L + 1], q[L])) L++;
			int j = q[L];
			dp[nw][i] = DP(i, j);
			from[k][i] = j;
			while (R - L > 1 && dY(i, q[R - 1]) * dX(q[R - 1], q[R - 2]) >= dY(q[R - 1], q[R - 2]) * dX(i, q[R - 1])) R--;
			q[R++] = i;
		}
		L(i, 1, N) dp[pv][i] = 0;
		L = R = 1;
	}
	i64 mx = -1;
	int start = -1;
	L(i, 1, N) {
		if (mx <= dp[N & 1][i]) {
			mx = dp[N & 1][i];
			start = i;
		}
	}
	cout << mx << '\n';
	L(i, 0, K - 1) {
		cout << start << ' ';
		start = from[K - i][start];
	}
}
/* 
contribution = sum in segment * sum after the segment

dp[i][j] = considered until i and have j splits
for all p < i
dp[i][j] = max(dp[p][j - 1] + (pre[i] - pre[p]) * (pre[N] - pre[i]))
dp[i][j] = max(dp[p][j - 1] + - pre[p] * (pre[N] - pre[i])) + (pre[i]) * (pre[N] - pre[i])

m = -pre[p]
x = pre[N] - pre[i]
c = dp[p][j - 1]

CHT:
	if x is better than y:
		dp[x][j - 1] - pre[x] * (pre[N] - pre[i]) >= dp[y][j - 1] - pre[y] * (pre[N] - pre[i])
		dp[x][j - 1] - dp[y][j - 1] >= (pre[x] - pre[y]) * (pre[N] - pre[i])
		
		Y(x) = dp[x]
		X(x) = pre[x]
		Y(x) - (y) >= (X(x) - X(y)) * (pre[N] - pre[i])
		slope(i, q[r]) <= slope(q[r], q[r - 1]) pop
*/

• Subtask 1: 0 / 11
• Subtask 2: 0 / 11
• Subtask 3: 0 / 11
• Subtask 4: 0 / 17
• Subtask 5: 0 / 21
• Subtask 6: 0 / 29