OUC_2014 Summer Training #9之Painting Fence

Description

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 a_i 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,?...,?a_n(1?≤?a_i?≤?10⁹).

Output

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

Sample Input

Input

5
2 2 1 2 1

Output

3

Input

2
2 2

Output

2

Input

1
5

Output

1

Hint

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.

费半天劲写了个程序，老开心了，一测试，上了趟厕所都没出结果，囧，定睛一看，满满的三重循环，囧。毕竟是花了时间写的，代码就保存一下吧，读者可以直接跳过。提醒：写程序前一定先考虑下时间的可行性。。。
#include
using namespace std;
int *plank;
int N;
class rectangle
{
public:
    rectangle()
    {
        a=0;
        b=0;
        head=0;
        tail=0;
    }
    int detect(int head,int n);
    void paint();
private:
    int a;   //宽
    int b;   //高
    int head;//头
    int tail;//尾
}rec[5000];


int main()
{
    int n,i,k,now,flag=0;
    cin>>n;
    plank=new int[n];
    for(i=0;i     {
        cin>>plank[i];
    }
    for(i=0;i     {
        if(plank[i]!=0)
        {
            flag=1;
            break;
        }
    }
    while(flag)
    {
        now=0;
        k=0;
        while(now!=n)
        {
            now=rec[k].detect(now,n);
            k++;
        }
        for(i=0;i         {
            rec[k].paint();
        }
        flag=0;
        for(i=0;i         {
            if(plank[i]!=0)
            {
                flag=1;
                break;
            }
        }
    }
    cout<     return 0;
}


int rectangle::detect(int now,int n)
{
    int i,tail,flag=0;
    a=0;
    b=0;
    for(i=now;i     {
        if(flag==0)
        {
            if(plank[i]!=0)
            {
                flag==1;
                head=i;
            }
        }
        if(flag==1)
        {
            if(plank[i]!=0)
                a++;
            else
            {
                break;
            }
        }
    }
    tail=i;
    for(i=head;i     {
        if(plank[i]>b)
            b=plank[i];
    }
    return tail;
}


void rectangle::paint()
{
    int i;
    if(b>a)
    {
        N+=a;
        for(i=head;i         {
            plank[i]=0;
        }
    }
    else
    {
        N+=1;
        for(i=head;i         {
            plank[i]--;
        }
    }
}

/*以下来自http://blog.csdn.net/u014733623/article/details/37927873*/

题意：你面前有宽度为1，高度给定的连续木板，每次可以刷一横排或一竖列，问你至少需要刷几次。

思路：先逆推，dp[i][j]表示第i列以后的木板都刷完了（不包括第i列）且前面的第j列是横着刷的（且假设它能够刷到后面的第i+1列），那么第i列之后不包括第i列最少需要的次数。所以当value[j]>=value[i]的时候第i列就不用刷了。

尽管这种做法会出现一些不合法的情况，但是最后的推出的结果运用到的必然都是合法的情况。

我们再正着看。

最后的答案应该是dp[0][0]，因为第0列是横着刷的就相当于没刷，第0列以后的都刷完了，就相当于刷完所有的。

它的来源是min(dp[1][0]+1,dp[1][1]+value[1])，即第一列是竖刷和第一列全为横刷的情况。剩下的依次类推。

AC代码如下：

#include
#include
#include
using namespace std;
int value[5010],dp[5010][5010];
int main()
{ int n,i,j,k;
  scanf("%d",&n);
  value[0]=0;
  for(i=1;i<=n;i++)
   scanf("%d",&value[i]);
  for(i=0;i<=n;i++)
   dp[n][i]=0;
  for(i=n;i>=1;i--)
   for(j=0;j
    if(value[j]>=value[i])
     dp[i-1][j]=dp[i][i];
    else
     dp[i-1][j]=min(dp[i][j]+1,dp[i][i]+value[i]-value[j]);
  printf("%d\n",dp[0][0]);
}