BZOJ2216 [Poi2011]Lightning Conductor 【决策单调性dp】

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;
}