[JSOI 2016] 灯塔

[题目链接]

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

[算法]

首先对不等式进行移项 :

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

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

显然 , sqrt(|i - j|)最多只有sqrt(n)个不同的值

用ST表求区间最值 ， 然后分块计算即可

时间复杂度： O(Nsqrt(N))

[代码]

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

int n;
int lg[MAXN] , bit[];
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 = ; x = ;
    char c = getchar();
    for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
    for (; isdigit(c); c = getchar()) x = (x << ) + (x << ) + c - '';
    x *= f;
}
inline long long query(int l,int r)
{
    int k = lg[r - l + ];
    return max(value[l][k],value[r - bit[k] + ][k]);
}

int main()
{

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

    return ;
}