[POI2011]Lightning Conductor
题目描述
Progressive climate change has forced the Byteburg authorities to build a huge lightning conductor that would protect all the buildings within the city.
These buildings form a row along a single street, and are numbered from 
to 
.
The heights of the buildings and the lightning conductor are non-negative integers.
Byteburg's limited funds allow construction of only a single lightning conductor.
Moreover, as you would expect, the higher it will be, the more expensive.
The lightning conductor of height 
located on the roof of the building 
(of height 
) protects the building 
(of height 
) if the following inequality holds:
where 
denotes the absolute value of the difference between 
and 
.
Byteasar, the mayor of Byteburg, asks your help.
Write a program that, for every building 
, determines the minimum height of a lightning conductor that would protect all the buildings if it were put on top of the building 
.
输入格式
In the first line of the standard input there is a single integer 
(
) that denotes the number of buildings in Byteburg.
Each of the following 
lines holds a single integer 
(
) that denotes the height of the 
-th building.
输出格式
Your program should print out exactly 
lines to the standard output.
The 
-th line should give a non-negative integer 
denoting the minimum height of the lightning conductor on the 
-th building.
题意翻译
给定一个长度为 nn 的序列 \{a_n\}{an}，对于每个 i\in [1,n]i∈[1,n] ，求出一个最小的非负整数 pp ，使得 \forall j\in[1,n]∀j∈[1,n]，都有 a_j\le a_i+p-\sqrt{|i-j|}aj≤ai+p−∣i−j∣
1 \le n \le 5\times 10^{5}1≤n≤5×105，0 \le a_i \le 10^{9}0≤ai≤109 。
输入输出样例
6 5 3 2 4 2 4
2 3 5 3 5 4
1 #include<iostream> 2 #include<cstdio> 3 #include<cmath> 4 using namespace std; 5 typedef long long ll; 6 int q[500005],R[500005],a[500005],n,b[500005]; 7 double sq[500005],f1[500005],f2[500005]; 8 double getValue(int x,int y) 9 { 10 return a[y]+(sq[x-y]); 11 } 12 int binary(int x,int y) 13 { 14 int l=x,r=n+1; 15 while (l<r) 16 { 17 int mid=(l+r)/2; 18 if (getValue(mid,y)<=getValue(mid,x)) r=mid; 19 else l=mid+1; 20 } 21 return l; 22 } 23 int main() 24 {int i,j; 25 scanf("%d",&n); 26 for (i=1;i<=n;i++) 27 { 28 scanf("%d",&a[i]); 29 sq[i]=sqrt((double)i); 30 } 31 int h=1,t=0; 32 for (i=1;i<=n;i++) 33 { 34 while (h<t&&R[t-1]>=binary(i,q[t])) t--; 35 R[t]=binary(i,q[t]); 36 q[++t]=i; 37 while (h<t&&R[h]<=i) h++; 38 f1[i]=getValue(i,q[h]); 39 } 40 h=1,t=0; 41 for (i=1;i<=n;i++) 42 b[i]=a[n-i+1]; 43 for (i=1;i<=n;i++) 44 a[i]=b[i]; 45 for (i=1;i<=n;i++) 46 { 47 while (h<t&&R[t-1]>=binary(i,q[t])) t--; 48 R[t]=binary(i,q[t]); 49 q[++t]=i; 50 while (h<t&&R[h]<=i) h++; 51 f2[n-i+1]=getValue(i,q[h]); 52 } 53 for (i=1;i<=n;i++) 54 printf("%.0lf %.0lf\n",f1[i],f2[i]); 55 for (i=1;i<=n;i++) 56 printf("%d\n",(int)ceil(max(f1[i],f2[i]))-a[n-i+1]); 57 }
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<cmath> 6 using namespace std; 7 int a[500005],n,b[500005]; 8 double sq[500005],f1[500005],f2[500005]; 9 void solve1(int l,int r,int L,int R) 10 {int i,Mid=0; 11 if (l>r) return; 12 int mid=(l+r)/2; 13 double s=0; 14 f1[mid]=a[mid];Mid=mid; 15 for (i=L;i<=min(R,mid);i++) 16 { 17 s=a[i]+sq[mid-i]; 18 if (s>f1[mid]) f1[mid]=s,Mid=i; 19 } 20 solve1(l,mid-1,L,Mid); 21 solve1(mid+1,r,Mid,R); 22 } 23 void solve2(int l,int r,int L,int R) 24 {int i,Mid=0; 25 if (l>r) return; 26 int mid=(l+r)/2; 27 double s=0; 28 f2[n-mid+1]=a[mid];Mid=mid; 29 for (i=L;i<=min(R,mid);i++) 30 { 31 s=a[i]+sq[mid-i]; 32 if (s>f2[n-mid+1]) f2[n-mid+1]=s,Mid=i; 33 } 34 solve2(l,mid-1,L,Mid); 35 solve2(mid+1,r,Mid,R); 36 } 37 int main() 38 {int i,j; 39 scanf("%d",&n); 40 for (i=1;i<=n;i++) 41 { 42 scanf("%d",&a[i]); 43 sq[i]=sqrt((double)i); 44 } 45 solve1(1,n,1,n); 46 for (i=1;i<=n;i++) 47 b[i]=a[n-i+1]; 48 for (i=1;i<=n;i++) 49 a[i]=b[i]; 50 solve2(1,n,1,n); 51 for (i=1;i<=n;i++) 52 printf("%d\n",(int)ceil(max(f1[i],f2[i]))-a[n-i+1]); 53 }
