CF448C Painting Fence(遞迴+貪心)

bigbigship發表於2014-07-20
AI遞迴
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
5
2 2 1 2 1
output
3
input
2
2 2
output
2
input
1
5
output
1
可以橫著放也可以豎著放 我們需要在其中選擇最優的 當橫著放的時候需要 minh（區間中最小的高度）次 豎著放需要res(區間中木板個數)次
每次選擇的為min(res,minh)  當高度有0的時候需要從此處分開 向兩邊進行同樣的操作
#include <iostream>
#include <algorithm>
#include <cstdio>
using namespace std;
const int N = 5001;
int n;
int a[N];
int solve(int l,int r)
{
   if(l>r)
      return 0;
   int minh=*min_element(a+l,a+r+1);//找到區間中最小元素的位置
   int ret=r-l+1,tot=minh;
   if(ret<minh){
      for(int i=l;i<=r;i++)
        a[i]=0;
      return ret;
   }
   else{
      for(int i=l;i<=r;i++)
        a[i]-=minh;
      int t=min_element(a+l,a+r+1)-a;
      tot+=(solve(l,t-1)+solve(t+1,r));
   }
   return min(ret,tot);
}
int main()
{
    while(cin>>n){
        for(int i=0;i<n;i++)
            cin>>a[i];
        cout<<solve(0,n-1)<<endl;
    }
    return 0;
}
/*****
5
2 2 1 2 1
***/










            




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