题目链接：

C. Painting Fence

time limit per test

1 second

memory limit per test

512 megabytes

input

standard input

output

standard output

Bizon the Champion isn't just attentive, he also is very hardworking.

Bizon the Champion decided to paint his old fence his favorite color, orange. The fence is represented as n vertical planks, put in a row. Adjacent planks have no gap between them. The planks are numbered from the left to the right starting from one, the i-th plank has the width of 1 meter and the height of ai meters.

Bizon the Champion bought a brush in the shop, the brush's width is 1 meter. He can make vertical and horizontal strokes with the brush. During a stroke the brush's full surface must touch the fence at all the time (see the samples for the better understanding). What minimum number of strokes should Bizon the Champion do to fully paint the fence? Note that you are allowed to paint the same area of the fence multiple times.

Input

The first line contains integer n (1 ≤ n ≤ 5000) — the number of fence planks. The second line contains n space-separated integersa1, a2, ..., an (1 ≤ ai ≤ 109).

Output

Print a single integer — the minimum number of strokes needed to paint the whole fence.

Sample test(s)

input

52 2 1 2 1

output

3

input

22 2

output

2

input

15

output

1

Note

In the first sample you need to paint the fence in three strokes with the brush: the first stroke goes on height 1 horizontally along all the planks. The second stroke goes on height 2 horizontally and paints the first and second planks and the third stroke (it can be horizontal and vertical) finishes painting the fourth plank.

In the second sample you can paint the fence with two strokes, either two horizontal or two vertical strokes.

In the third sample there is only one plank that can be painted using a single vertical stroke.

题意：

给篱笆上色。要求步骤最少，篱笆怎么上色应该懂吧。，。刷子能够在横着和竖着刷，不能跳着刷。

假设是竖着刷，应当是篱笆的条数。横着刷的话。就是刷完最短木板的长度。再接着考虑没有刷的木板中最短的。

代码例如以下：

#include 
     
      #include 
      
       using namespace std;#define INF 0x3fffffffint n, a[5017];int dfs(int sl, int sr){	int MIN = INF, num = 0;	if(sl > sr)	return 0;	for(int i = sl; i <= sr; i++)	{		if(a[i] < MIN)		{			MIN = a[i];		}	}	for(int i = sl; i <= sr; i++)	{		a[i]-=MIN;	}	num+=MIN;	int ll = sl;	for(int i = sl; i <= sr; i++)	{		if(a[i] == 0)//假设a[i]=0,中断处 		{			num+=dfs(ll,i-1);			ll =i+1;		}	}		if(ll <= sr)//最后一根不是零的情况 		{			num+=dfs(ll,sr);		}	return min(num,sr-sl+1);}int main(){	int i, j;	while(cin>>n)	{		for(i = 1; i <= n; i++)		{			cin>>a[i];		}		int ans = dfs(1,n);		cout<
       
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