D. Bear and Blocks
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
Limak is a little bear who loves to play. Today he is playing by destroying block towers. He built n towers in a row. The i-th tower is made of hi identical blocks. For clarification see picture for the first sample.
Limak will repeat the following operation till everything is destroyed.
Block is called internal if it has all four neighbors, i.e. it has each side (top, left, down and right) adjacent to other block or to the floor. Otherwise, block is boundary. In one operation Limak destroys all boundary blocks. His paws are very fast and he destroys all those blocks at the same time.
Limak is ready to start. You task is to count how many operations will it take him to destroy all towers.
Input
The first line contains single integer n (1 ≤ n ≤ 105).
The second line contains n space-separated integers h1, h2, …, hn (1 ≤ hi ≤ 109) — sizes of towers.
Output
Print the number of operations needed to destroy all towers.
Examples
inputCopy
6
2 1 4 6 2 2
outputCopy
3
inputCopy
7
3 3 3 1 3 3 3
outputCopy
2
超水的d题，不知道为什么才这么点人过。。。
第i列方格消失的时间取决于三个方面：i列的高度，左边一列完全消失的时间，右边一列完全消失的时间
状态转移方程即：dp[i]=min(dp[i-1]+1,dp[i+1]+1,h[i])
/* ***********************************************
Author :ACagain
Created Time :2018/4/13 19:08:48
File Name :4_13.cpp
************************************************ */
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <stack>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <queue>
#include <set>
#include <map>
using namespace std;
#define lson o<<1,l,m
#define rson o<<1|1,m+1,r
#define pii pair<int,int>
#define mp make_pair
#define ll long long
#define INF 0x3f3f3f3f
const int maxn=1e5+5;
int h[maxn],dp[maxn];
int main()
{
std::ios::sync_with_stdio(false);
std::cin.tie(0);
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
int n;
cin>>n;
for(int i=1;i<=n;i++)
cin>>h[i];
dp[0]=dp[n+1]=0;
for(int i=1;i<=n;i++)
{
dp[i]=min(h[i],dp[i-1]+1);
}
int ans=0;
for(int i=n;i>=1;i--)
{
dp[i]=min(dp[i],dp[i+1]+1);
ans=max(dp[i],ans);
}
cout<<ans<<endl;
return 0;
}