递归解Codeforces Round #256 (Div. 2)C. Painting Fence

#include<iostream>
#include<map>
#include<string>
#include<cstring>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<queue>
#include<vector>
#include<algorithm>
using namespace std;
int n,a[5010];
int dfs(int l,int r,int sub)
{
	if(l>r)
		return 0;
	int t=min_element(a+l,a+r+1)-a;//获取这个区间内最小的元素下标
	return min(a[t]+dfs(l,t-1,a[t])+dfs(t+1,r,a[t])-sub,r-l+1);
	//最小次数等于，横着刷导致的之后刷的最小次数之和，竖着刷需要的次数，它们俩的最小值
}
int main()
{
	cin>>n;
	for(int i=0;i<n;i++)
		cin>>a[i];
	cout<<dfs(0,n-1,0);
}

递归解Codeforces Round #256 (Div. 2)C. Painting Fence