题目链接
BZOJ2216
题解
学过高中数学都应知道，我们要求\(p\)的极值，参变分离为
\[h_j + sqrt{|i - j|} - h_i \le p\]
实际上就是求\(h_j + sqrt{|i - j|} - h_i\)的最大值
就可以设\(f[i]\)表示对\(i\)最大的该式的值
绝对值通常要去掉，一般可以通过方向性，我们只需每次转移时令\(i > j\)，正反转移两次即可
现在式子变为
\[f[i] = max\{h_j + \sqrt{i - j}\} - h_i\]
发现\(\sqrt{i - j}\)依旧无法处理，无法展开使用我们喜闻乐见的斜率优化
此时就可以考虑这个式子是否具有决策单调性
我们考虑对于\(i'<i\)，我们的决策为\(h_j + sqrt{i' - j}\)
那么对于\(forall k < j\)有\(h_k + sqrt{i' - k} < h_j + sqrt{i' - j}\)
现在我们用\(i\)替换\(i'\)
式子变为\(h_k + sqrt{i - k}\)和\(h_j + sqrt{i - j}\)
\(h_k\)和\(h_j\)是没有变化的，如果\(sqrt{i - j}\)的增长比\(sqrt{i - k}\)的增长要快，我们就可认定\(i\)替换\(i'\)后，\(k\)依旧无法作为最优决策
考虑函数
\[f(x) = \sqrt{x}\]
\[f'(x) = \frac{1}{2\sqrt{x}}\]
显然当\(x\)越大增长率越慢，而\(i' - k > i' - j\)，\(\sqrt{i - j}\)的增长的确比\(\sqrt{i - k}\)的增长要快
得证
所以用队列维护三元组优化即可
复杂度\(O(nlogn)\)
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<map>
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define mp(a,b) make_pair<int,int>(a,b)
#define cls(s) memset(s,0,sizeof(s))
#define cp pair<int,int>
#define LL long long int
using namespace std;
const int maxn = 500005,maxm = 100005;
inline int read(){int out = 0,flag = 1; char c = getchar();while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}return out * flag;
}
double f[maxn],h[maxn];
int n,head,tail,ans[maxn];
struct tri{int l,r,pos;}q[maxn << 1];
inline double cal(int i,int j){return h[j] + sqrt(i - j) - h[i];
}
inline bool check(int pos,int i,int j){return cal(pos,i) >= cal(pos,j);
}
void work(){q[head = tail = 0] = (tri){1,n,1};tri u;for (int i = 1; i <= n; i++){ans[i] = max(ans[i],(int)ceil(cal(i,q[head].pos)));q[head].l++;if (q[head].l > q[head].r) head++;while (head <= tail){u = q[tail--];if (!check(u.r,i,u.pos)){q[++tail] = u;if (u.r + 1 <= n) q[++tail] = (tri){u.r + 1,n,i};break;}if (check(u.l,i,u.pos)){if (head > tail){q[++tail] = (tri){i + 1,n,i};break;}continue;}else {int l = u.l,r = u.r,mid;while (l < r){mid = l + r >> 1;if (check(mid,i,u.pos)) r = mid;else l = mid + 1;}q[++tail] = (tri){u.l,l - 1,u.pos};q[++tail] = (tri){l,n,i};break;}}}
}
int main(){n = read();for (int i = 1; i <= n; i++) h[i] = read();work();reverse(h + 1,h + 1 + n);reverse(ans + 1,ans + 1 + n);work();for (int i = n; i; i--)printf("%d\n",ans[i]);return 0;
}
