cf #256 (Div. 2) C. Painting Fence 分治

最新推荐文章于 2023-04-25 10:22:54 发布

luckyone2014

最新推荐文章于 2023-04-25 10:22:54 发布

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分类专栏： dfs 文章标签： cf 算法分治

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本文深入探讨了如何通过分治策略解决C. Painting Fence问题，即计算最少涂漆次数以完全覆盖长条形栅栏。通过分析输入数据，作者提出了将问题分解为子问题的方法，并计算每个子问题的最优解，最终得出整体最优解。实例演示了该算法的应用过程，详细解释了计算步骤和关键决策，使读者能够理解并应用到类似问题中。

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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

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.

渣渣昨天被二分玩弄了一天, 精疲力竭.所以没去做C .感觉C 不是很难.

我用分治做的,看到别人还有用dp的,dp做法自行百度吧.

我这个是想着判断每次去掉最短边, 然后判断是横着画比较快,还是直接把这段区间竖着直接涂满的快;

#include<stdio.h>
#include<algorithm>
#include<set>
using namespace std;
 
#define IN freopen("input.txt", "r", stdin)    
#define OUT freopen("output.txt", "w", stdout)
int a[10000];
int min(int a,int b)
{
	return a<b?a:b;
}
int solve(int l,int r)
{
	int wide=r-l+1;
	if(wide<=0)
		return 0;
	if(wide==1)
		return 1;
	int minh=a[l],i,times;
	for(i=l+1;i<=r;i++)
	{
		minh=min(minh,a[i]);
	}
	for(i=l;i<=r;i++)
		a[i]-=minh;//减去最短;
    times=minh;
	
	
	for(i=l;i<=r;i++)
	{
    	while(a[i]==0&&i<=r) i++;
		 int fro=i;
		while(a[i]!=0&&i<=r) i++;
		 int rea=i-1;
		times+=solve(fro,rea);//涂掉最短边后,子区间计算出最优解.
	}
    return min(wide,times);  //wide 是这段区间的宽,表示竖着涂用时,times横着涂完最小边之后的最优解.
}
int main()
{ 
	//IN;
	//OUT;

	int i,n;
	while(scanf("%d",&n)!=EOF)
	{
		for(i=1;i<=n;i++)
	    	scanf("%d",&a[i]);
		printf("%d\n",solve(1,n));
	}
		
	return 0;
}