bzoj2216: [Poi2011]Lightning Conductor

Description

已知一个长度为n的序列a1,a2,...,an。
对于每个1<=i<=n，找到最小的非负整数p满足 对于任意的j, aj < = ai + p - sqrt(abs(i-j))

Input

第一行n，(1<=n<=500000)
下面每行一个整数，其中第i行是ai。(0<=ai<=1000000000)

 

Output

n行，第i行表示对于i，得到的p

 

Sample Input

6
5
3
2
4
2
4

Sample Output

2
3
5
3
5
4

 

题解：

http://hzwer.com/5273.html

code：

#include<cstdio>
#include<iostream>
#include<cmath>
#include<cassert>
#include<cstring>
#include<algorithm>
using namespace std;
char ch;
bool ok;
void read(int &x){
    for (ok=0,ch=getchar();!isdigit(ch);ch=getchar()) if (ch=='-') ok=1;
    for (x=0;isdigit(ch);x=x*10+ch-'0',ch=getchar());
    if (ok) x=-x;
}
const int maxn=500005;
int n,a[maxn],ans[maxn],root[maxn];
double calc(int j,int i){return a[j]-a[i]+sqrt(i-j);}
struct Stack{
    int top,pos;
    struct Data{
        int l,r,id;
    }s[maxn],tmp;
    void init(){s[top=1]=(Data){2,n,1},pos=1;}
    bool cmp(int a,int x,int y){return calc(x,a)>calc(y,a);}
    int get(int id){
        int l=tmp.l,r=tmp.r,m,a=tmp.id;
        while (l<r){
            m=((l+r)>>1)+1;
            if (cmp(m,a,id)) l=m; else r=m-1;
        }
        return l;
    }
    void push(int id){
        while (top>=pos&&!cmp(s[top].l,s[top].id,id)) top--;
        if (top>=pos){
            tmp=s[top--];
            int m=get(id);
            if (tmp.l<=m) s[++top]=(Data){tmp.l,m,tmp.id};
            if (m<n) s[++top]=(Data){m+1,n,id};
        }
        else s[++top]=(Data){id+1,n,id};
    }
    int query(int x){
        while (x>s[pos].r) pos++;
        s[pos].l=x+1;
        return ceil(calc(s[pos].id,x));
    }
}stack;
int main(){
    read(n);
    for (int i=1;i<=n;i++) read(a[i]);
    stack.init();
    for (int i=2;i<=n;i++) ans[i]=max(0,stack.query(i)),stack.push(i);
    for (int i=1,j=n;i<j;i++,j--) swap(a[i],a[j]);
    stack.init();
    for (int i=2;i<=n;i++) ans[n-i+1]=max(ans[n-i+1],stack.query(i)),stack.push(i);
    for (int i=1;i<=n;i++) printf("%d\n",ans[i]);
    return 0;
}