codeforces 574D. Bear and Blocks

最新推荐文章于 2019-05-30 15:19:49 发布

Kirito_Acmer

最新推荐文章于 2019-05-30 15:19:49 发布

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本文解析了一道关于熊摧毁由方块堆叠而成的塔楼的算法题目。通过定义内部与边界方块的区别，介绍了如何使用动态规划的方法来计算熊完全摧毁所有塔楼所需的最少操作次数。

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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.

Sample test(s)

input
6
2 1 4 6 2 2

output
3

input
7
3 3 3 1 3 3 3

output
2

Note

The picture below shows all three operations for the first sample test. Each time boundary blocks are marked with red color.

$\text{[math]}$

After first operation there are four blocks left and only one remains after second operation. This last block is destroyed in third operation.

刚用模拟做T了，后来发现可以用简单的dp做。记录l[i]为从左到右消去这一行的最小操作数，r[i]为从右到左消去这一行的最小操作数，那么l[i]=r[n]=1.

l[i]=min(h[i],l[i-1]+1),r[i]=min(h[i],r[i+1]+1)，然后只要求出最大操作数t[i]=max(l[i],r[i])的最小值就行了。

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<string>
#include<algorithm>
using namespace std;
#define ll long long
#define inf 0x7fffffff
#define maxn 100060
int l[maxn],r[maxn],h[maxn],t[maxn];

int main()
{
    int n,m,i,j,ans;
    while(scanf("%d",&n)!=EOF)
    {
        for(i=1;i<=n;i++){
            scanf("%d",&h[i]);
        }
        l[1]=1;r[n]=1;
        for(i=2;i<=n;i++){
            l[i]=min(h[i],l[i-1]+1);

        }
        for(i=n-1;i>=1;i--){
            r[i]=min(h[i],r[i+1]+1);
        }
        ans=0;
        for(i=1;i<=n;i++){
            t[i]=min(l[i],r[i]);
            ans=max(t[i],ans);
        }
        printf("%d\n",ans);

    }
    return 0;
}