AtCoder Regular Contest 066 F Contest with Drinks Hard

洛谷传送门

AtCoder 传送门

下文令 $a$ 为原题中的 $T$。

考虑若没有饮料，可以设 $f_i$ 表示，考虑了前 $i$ 道题，第 $i$ 道题没做的最大得分。转移就枚举上一道没做的题 $j$，那么 $[j + 1, i - 1]$ 形成一个连续段。设 $b_i$ 为 $a_i$ 的前缀和，可得：

$$f_i = \max\limits_{j = 0}^{i - 1} \{ f_j + \frac{(i - j)(i - j - 1)}{2} - b_{i - 1} + b_j \}$$

拆项可得：

$$f_i - \frac{i(i - 1)}{2} + b_{i - 1} = \max\limits_{j = 0}^{i - 1} \{ f_j + \frac{j(j + 1)}{2} + b_j - ij \}$$

容易发现其为斜率优化形式，$f_i - \frac{i(i - 1)}{2} + b_{i - 1}$ 是截距，$f_j + \frac{j(j + 1)}{2} + b_j$ 是纵坐标，$i$ 是斜率，$j$ 是横坐标。注意到斜率和横坐标都单增，所以需要使用单调栈维护上凸包。

考虑若有饮料，设饮料把 $a_x$ 变成 $y$。分类讨论是否使用饮料。再求一个 $g_i$ 表示考虑了第 $i$ 道题到第 $n$ 道题，第 $i$ 道题没做的最大得分。

若没使用饮料，答案是 $f_x + g_x$；

若使用饮料，答案是 $h_x + a_x - y$。其中 $h_i$ 表示必须做第 $i$ 道题的最大得分。

考虑如何求 $h_i$。有一个单次 $O(n^2)$ 的求法：

$$h_i = \max\limits_{l < i < r} \{ f_l + g_r + \frac{(r - l)(r - l - 1)}{2} - b_{r - 1} + b_l \}$$

套路地，考虑分治。设当前分治区间为 $[L, R]$，$mid = \left\lfloor\frac{L + R}{2}\right\rfloor$。考虑 $l \in [L, mid], r \in [mid + 1, R]$，对 $i \in (L, R)$ 造成的贡献。分类讨论 $i$ 在哪半边，例如若 $i \in (L, mid], l \in [L, i - 1], r \in [mid + 1, R]$，就求一个 $[mid + 1, R]$ 的凸包，然后遍历 $l$，双指针遍历凸包即可。

时间复杂度为 $O(n \log n)$。

code

// Problem: F - Contest with Drinks Hard
// Contest: AtCoder - AtCoder Regular Contest 066
// URL: https://atcoder.jp/contests/arc066/tasks/arc066_d
// Memory Limit: 256 MB
// Time Limit: 2000 ms
// 
// Powered by CP Editor (https://cpeditor.org)
 
#include <bits/stdc++.h>
#define pb emplace_back
#define fst first
#define scd second
#define mkp make_pair
#define mems(a, x) memset((a), (x), sizeof(a))
 
using namespace std;
typedef long long ll;
typedef double db;
typedef unsigned long long ull;
typedef long double ldb;
typedef pair<ll, ll> pii;
 
const int maxn = 300100;
 
ll n, m, a[maxn], b[maxn], f[maxn], g[maxn], h[maxn], stk[maxn], top;
struct node {
	ll x, y;
	node(ll a = 0, ll b = 0) : x(a), y(b) {}
} c[maxn];
 
inline node operator + (const node &a, const node &b) {
	return node(a.x + b.x, a.y + b.y);
}
 
inline node operator - (const node &a, const node &b) {
	return node(a.x - b.x, a.y - b.y);
}
 
inline ll operator * (const node &a, const node &b) {
	return a.x * b.y - a.y * b.x;
}
 
void dfs(int l, int r) {
	if (l == r) {
		return;
	}
	int mid = (l + r) >> 1, tot = 0, top = 0;
	for (int i = mid + 1; i <= r; ++i) {
		c[++tot] = node(i, g[i] + 1LL * i * (i - 1) / 2 - b[i - 1]);
	}
	for (int i = 1; i <= tot; ++i) {
		while (top > 1 && (c[i] - c[stk[top - 1]]) * (c[stk[top]] - c[stk[top - 1]]) <= 0) {
			--top;
		}
		stk[++top] = i;
	}
	ll mx = -1e18;
	// printf("l, r: %d %d %d\n", l, mid, r);
	for (int i = l, j = top; i <= mid; ++i) {
		while (j > 1 && (c[stk[j]].y - c[stk[j - 1]].y) <= (c[stk[j]].x - c[stk[j - 1]].x) * i) {
			--j;
		}
		int k = stk[j] + mid;
		h[i] = max(h[i], mx);
		// printf("%d %d %lld\n", i, k, f[i] + g[k] + 1LL * (k - i) * (k - i - 1) / 2 - b[k - 1] + b[i]);
		mx = max(mx, f[i] + g[k] + 1LL * (k - i) * (k - i - 1) / 2 - b[k - 1] + b[i]);
	}
	tot = top = 0;
	for (int i = l; i <= mid; ++i) {
		c[++tot] = node(i, f[i] + 1LL * i * (i + 1) / 2 + b[i]);
	}
	for (int i = 1; i <= tot; ++i) {
		while (top > 1 && (c[i] - c[stk[top - 1]]) * (c[stk[top]] - c[stk[top - 1]]) <= 0) {
			--top;
		}
		stk[++top] = i;
	}
	mx = -1e18;
	for (int i = r, j = 1; i > mid; --i) {
		while (j < top && (c[stk[j + 1]].y - c[stk[j]].y) >= (c[stk[j + 1]].x - c[stk[j]].x) * i) {
			++j;
		}
		int k = stk[j] + l - 1;
		h[i] = max(h[i], mx);
		mx = max(mx, f[k] + g[i] + 1LL * (i - k) * (i - k - 1) / 2 - b[i - 1] + b[k]);
	}
	dfs(l, mid);
	dfs(mid + 1, r);
}
 
void solve() {
	scanf("%lld", &n);
	for (int i = 1; i <= n; ++i) {
		scanf("%lld", &a[i]);
		b[i] = b[i - 1] + a[i];
	}
	scanf("%lld", &m);
	stk[++top] = 0;
	for (int i = 1; i <= n; ++i) {
		while (top > 1 && (c[stk[top]].y - c[stk[top - 1]].y) <= (c[stk[top]].x - c[stk[top - 1]].x) * i) {
			--top;
		}
		int j = stk[top];
		f[i] = f[j] + 1LL * (i - j) * (i - j - 1) / 2 - b[i - 1] + b[j];
		c[i] = node(i, f[i] + 1LL * i * (i + 1) / 2 + b[i]);
		while (top > 1 && (c[i] - c[stk[top - 1]]) * (c[stk[top]] - c[stk[top - 1]]) <= 0) {
			--top;
		}
		stk[++top] = i;
	}
	reverse(a + 1, a + n + 1);
	for (int i = 1; i <= n; ++i) {
		b[i] = b[i - 1] + a[i];
	}
	mems(stk, 0);
	top = 0;
	stk[++top] = 0;
	for (int i = 1; i <= n; ++i) {
		while (top > 1 && (c[stk[top]].y - c[stk[top - 1]].y) <= (c[stk[top]].x - c[stk[top - 1]].x) * i) {
			--top;
		}
		int j = stk[top];
		g[i] = g[j] + 1LL * (i - j) * (i - j - 1) / 2 - b[i - 1] + b[j];
		c[i] = node(i, g[i] + 1LL * i * (i + 1) / 2 + b[i]);
		while (top > 1 && (c[i] - c[stk[top - 1]]) * (c[stk[top]] - c[stk[top - 1]]) <= 0) {
			--top;
		}
		stk[++top] = i;
	}
	reverse(a + 1, a + n + 1);
	reverse(g + 1, g + n + 1);
	for (int i = 1; i <= n; ++i) {
		b[i] = b[i - 1] + a[i];
	}
	mems(h, -0x3f);
	dfs(0, n + 1);
	// for (int i = 1; i <= n; ++i) {
		// printf("h[%d] = %lld\n", i, h[i]);
	// }
	// for (int i = 1; i <= n + 1; ++i) {
		// printf("g[%d] = %lld\n", i, g[i]);
	// }
	while (m--) {
		ll x, y;
		scanf("%lld%lld", &x, &y);
		printf("%lld\n", max(f[x] + g[x], h[x] + a[x] - y));
	}
}
 
int main() {
	int T = 1;
	// scanf("%d", &T);
	while (T--) {
		solve();
	}
	return 0;
}