[題目鏈接]

https://www.lydsy.com/JudgeOnline/problem.php?id=4850

[算法]

首先對不等式進行移項 :

hj <= hi + p - sqrt(|i - j|)

p >= hj - hi + sqrt(|i - j|)

顯然 , sqrt(|i - j|)最多只有sqrt(n)個不同的值

用ST表求區間最值 ， 然後分塊計算即可

時間復雜度： O(Nsqrt(N))

[代碼]

#include<bits/stdc++.h>
using
 
 namespace std;
#define MAXN 200010
#define MAXLOG 20
#define sqr(x) x * x

int n;
int lg[MAXN] , bit[25];
long long h[MAXN];
long long value[MAXN][MAXLOG];

template <typename T> inline void chkmax(T &x,T y) { x = max(x,y); }
template <typename T> inline void read(T &x)
{
    T f = 1
 
; x = 0;
    char c = getchar();
    for (; !isdigit(c); c = getchar()) if (c == ‘-‘) f = -f;
    for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - ‘0‘;
    x *= f;
}
inline long long query(int l,int r)
{
    int k = lg[r - l + 1];
    return max(value[l][k],value[r - bit[k] + 1
 
][k]);
}

int main()
{
    
    read(n);
    for (register int i = 1; i < MAXN; i++) lg[i] = (double)log(i) / log(2.0);
    bit[0] = 1;
    for (register int i = 1; i <= 20; i++) bit[i] = bit[i - 1] << 1;
    for (register int i = 1; i <= n; i++) read(h[i]);
    for (register int i = 1; i <= n; i++) value[i][0] = h[i];
    for (register int i = 1; i < MAXLOG; i++)
    {
        for (register int j = 1; j + (1 << i) <= n; j++)
        {
            value[j][i] = max(value[j][i - 1],value[j + bit[i - 1]][i - 1]);
        }    
    }    
    for (register int i = 1; i <= n; i++)
    {
        int l = i , r , sq = 1;
        long long ans = 0;
        while (l != 1)
        {                  
            r = l - 1;
            l = max(1,i - sqr(sq));
            chkmax(ans,sq + query(l,r) - h[i]);
            sq++;
        }    
        r = i , sq = 1;
        while (r != n)
        {
            l = r + 1;
            r = min(n,i + sqr(sq));
            chkmax(ans,sq + query(l,r) - h[i]);    
            sq++;        
        }
        printf("%lld\n",ans);
    }
    
    return 0;
}

[JSOI 2016] 燈塔