BZOJ 4850 [Jsoi2016]灯塔

题解:

发现根号里面只有根号种取值

枚举相同的取值，rmq计算最值即可

rmq取log要提前预处理

//RMQ长度预处理 
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#define max(a,b) ((a>b)?a:b)
using namespace std;
const int maxn=100009;
const int oo=1000000000;

int n;
int h[maxn];
int lgs[maxn];

int f[maxn][20];
void STinit(){
    for(int i=1;i<=n;++i)f[i][0]=h[i];
    for(int j=1;j<=19;++j){
        for(int i=1;i+(1<<j)-1<=n;++i){
            f[i][j]=max(f[i][j-1],f[i+(1<<(j-1))][j-1]);
        }
    }
}
inline int Querymax(int l,int r){
    int k=lgs[r-l+1];
    return max(f[l][k],f[r-(1<<k)+1][k]);
}

int rd(){
    int r=0,k=1;
    char c=getchar();
    for(;c<'0'||c>'9';c=getchar())if(c=='-')k=-1;
    for(;c>='0'&&c<='9';c=getchar())r=r*10+c-'0';
    return r*k;
}
int main(){
//    freopen("light.in","r",stdin);
//    freopen("light.out","w",stdout);
    
    scanf("%d",&n);
    for(int i=1;i<=n;++i)h[i]=rd();
    STinit();
    for(int i=1;i<=n;++i)lgs[i]=(int)log2(i+0.5);
    
    for(int i=1;i<=n;++i){
        int ans=0;
        for(int j=1;j<=n;++j){
            int fla=0;
            int r=i-(j-1)*(j-1)-1;
            int l=i-j*j;
            if(r<=0){
                fla=1;
            }else{
                l=max(1,l);
                ans=max(ans,-h[i]+j+Querymax(l,r));
            }

            l=i+(j-1)*(j-1)+1;
            r=i+j*j;
            if(l>n){
                fla&=1;
            }else{
                fla=0;
                r=min(r,n);
                ans=max(ans,-h[i]+j+Querymax(l,r));
            }
            if(fla)break;
        }
        printf("%d
",ans);
    }
    return 0;
}