题目

题解

对于 \(m=1\) 这档分
我们可以 \(dp\) 然后斜率优化
具体来说就是 \(f_i = f_j + \frac{(i-j)\times (i-j+1)}{2} + sum[j]-sum[i]\)
很容易斜率优化

那么 \(m=3\times 10^5\) 时
考虑 \(cdq\) 分治
考虑 \(dp\) 出一个如上定义的前缀 \(f\) 和同理的后缀 \(g\)
设 \(h_x\) 表示强制选 \(x\) 时的最大收益，\(h_x = f_i+g_j-sum_{j-1}+sum_i+\frac{(j-i)\times(j-i-1)}{2}(i < x <j)\)
这个 \(O(n^2)\) 显然不行
记 \(F_i = f_i+\frac{i \times (i-1)}{2}+sum_i，G_i = g_i + \frac{i \times (i-1)}{2}-sum_{i-1}\)
则 \(h_x = F_i + G_j - i \times j(i < x < j)\)
对于区间 \([l，r]\)，我们讨论 \([l，mid]\) 的 \(h\) 值
对于一个 \(l\le x\le mid\)，它的 \(h\) 值 \(i\) 决策在它左边，\(j\) 决策 \(mid\) 右边
我们考虑对 \(j\) 决策斜率优化，\(i\) 递增，和 \(f,g\) 的算法有异曲同工之妙

然后通过一些玄学操作同理处理决策 \(i\)
最后答案就是 \(\max(f_{p-1}+g_{p+1}，h_p+a_p-x)\)

\(Code\)

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#define LL long long
using namespace std;

const int N = 3e5 + 5;
const LL INF = 0x3f3f3f3f3f3f3f3f;
int n, m, a[N], Q[N];
LL sum[N], f[N], g[N], F[N], G[N], h[N], w[N];

inline double slope(LL *f, int j, int k)
{
	return 1.0 * ((f[j] + (1LL * j * j - j) / 2.0 + sum[j]) - (f[k] + (1LL * k * k - k) / 2.0 + sum[k])) / (j - k);
}
inline void dp(LL *f)
{
	memset(f, 0, sizeof f);
	int top;
	Q[top = 1] = 0;
	for(register int i = 1; i <= n; i++)
	{
		while (top > 1 && slope(f, Q[top - 1], Q[top]) < i) --top;
		f[i] = max(f[i - 1], f[Q[top]] + 1LL * (i - Q[top]) * (i - Q[top] + 1) / 2 - sum[i] + sum[Q[top]]);
		while (top && slope(f, Q[top - 1], Q[top]) < slope(f, Q[top], i)) --top;
		Q[++top] = i;
	}
}

inline double slope(int j, int k){return 1.0 * (G[j] - G[k]) / (j - k);}
void divide(int l, int r, int p)
{
	int mid = (l + r) >> 1, top = 0;
	for(register int i = mid + 1; i <= r; i++)
	{
		while (top > 1 && slope(Q[top - 1], Q[top]) < slope(Q[top], i)) --top;
		Q[++top] = i;
	}
	w[l - 1] = -INF;
	for(register int i = l; i <= mid; i++)
	{
		while (top > 1 && slope(Q[top - 1], Q[top]) < i) --top;
		w[i] = max(w[i - 1], F[i] + G[Q[top]] - 1LL * i * Q[top]);
	}
	if (p) for(register int i = l; i <= mid; i++) h[n - i + 1] = max(h[n - i + 1], w[i - 1]);
	else for(register int i = l; i <= mid; i++) h[i] = max(h[i], w[i - 1]);
	if (l ^ r) divide(l, mid, p), divide(mid + 1, r, p);
}

void solve()
{
	for(register int i = 1; i <= n; i++) sum[i] = sum[i - 1] + a[i];
	dp(f);
	reverse(a + 1, a + n + 1);
	for(register int i = 1; i <= n; i++) sum[i] = sum[i - 1] + a[i];
	dp(g);
	
	reverse(f + 1, f + n + 1);
	for(register int i = 1; i <= n; i++) swap(f[i], g[i]);
	for(register int i = 0; i <= n + 1; i++) F[i] = f[i] + 1LL * i * (i + 1) / 2 + sum[i], G[i] = g[i] + 1LL * i * (i - 1) / 2 - sum[i - 1];
	for(register int i = 1; i <= n; i++) h[i] = -INF;
	divide(0, n + 1, 1);
	
	for(register int i = 1; i <= n; i++) swap(f[i], g[i]);
	reverse(a + 1, a + n + 1), reverse(f + 1, f + n + 1), reverse(g + 1, g + n + 1);
	for(register int i = 1; i <= n; i++) sum[i] = sum[i - 1] + a[i];
	for(register int i = 0; i <= n + 1; i++) F[i] = f[i] + 1LL * i * (i + 1) / 2 + sum[i], G[i] = g[i] + 1LL * i * (i - 1) / 2 - sum[i - 1];
	divide(0, n + 1, 0);
}

int main()
{
	freopen("genocide.in", "r", stdin);
	freopen("genocide.out", "w", stdout);
	scanf("%d", &n);
	for(register int i = 1; i <= n; i++) scanf("%d", &a[i]);
	solve();
	scanf("%d", &m);
	for(int p, x; m; --m)
	{
		scanf("%d%d", &p, &x);
		printf("%lld\n", max(f[p - 1] + g[p + 1], h[p] + a[p] - x));
	}
}