**題意：**n個寬度為1，高度為ai的板子，橫豎一筆刷漆，問最小需要多少筆刷完。
思路：貪心加分治，首先對於一些[l, r]序號區間內的板子，豎著刷為(r-l+1)花費，橫著刷的話肯定要先刷到最小的長度的那個，然後分成左右兩半，然後分治重複遞迴。

#include<iostream>
#include<string>
#include<cstdio>
#include<cstring>
#include<bitset>
#include<algorithm>
#include<map>
#include<set>
 

#include<queue>
#include<vector>
#include<cstdlib>
#include<list>
#include<stack>
#include<cmath>
#include<iomanip>
using namespace std;
//#pragma comment(linker, "/STACK:1024000000,1024000000")
typedef long long LL;
void debug() {cout << "ok running!" << endl;}
int
 
 n, a[5005];
int solve(int l, int r)
{
    if(l == r) return a[l]!=0;
    if(l > r) return 0;
    //cout << l << " " << r << endl;
    int m = 0x3f3f3f3f, pos = -1;
    int res = (r-l+1);
    for(int i = l; i <= r; ++i)
    {
        if(a[i] < m)
        {
            m = a[i];
            pos = i;
        }
    }
    for
 
(int i = l; i <= r; ++i)
        a[i] -= m;

    return min(res, solve(l, pos-1) + solve(pos+1, r) + m);
}
int main()
{
    ios::sync_with_stdio(false);
    #ifndef ONLINE_JUDGE
    //freopen("input.txt", "r", stdin);
    #endif // ONLINE_JUDGE
    while(cin >> n)
    {
        for(int i = 0; i < n; ++i)
        {
            cin >> a[i];
        }
        int ans = solve(0, n-1);
        cout << ans << endl;
    }
    return 0;
}