Painting Fence

最新推荐文章于 2022-05-08 17:24:11 发布

转载最新推荐文章于 2022-05-08 17:24:11 发布 · 164 阅读

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原文链接：http://www.cnblogs.com/superxuezhazha/p/5693036.html

本文探讨了一个有趣的涂色问题：如何用最少的笔划将不同高度的围栏板全部涂成橙色。通过递归算法，对比了全部横涂与竖涂所需的最小次数，并给出了完整的代码实现。

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Painting Fence

Time Limit:1000MS Memory Limit:524288KB 64bit IO Format:%I64d & %I64u

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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

Input

2
2 2

Output

Input

1
5

Output

#include<cstdio>  
#include<algorithm>  
using namespace std;  
const int MAXN = 5005;  
int a[MAXN];  
int solve(int l, int r)  
{  
    int k = l;  
    if(l > r) return 0;  
    for(int i = l; i <= r; i++)  
        if(a[i] < a[k])  
            k = i;  
    int tmp = a[k];  
    for(int i = l; i <= r; i++)  
        a[i] -= tmp;  
    return min(r-l+1, solve(l, k-1) + solve(k+1, r) + tmp);  
}  
int main()  
{  
    int n;  
    scanf("%d",&n);  
    for(int i = 0; i < n; i++)  
        scanf("%d",&a[i]);  
    printf("%d\n", solve(0, n-1));  
    return 0;  
}

View Code

递归，每次做的到操作都是比较全部横着涂完和全部竖着涂完的涂刷到次数较小的次数，递归出答案。

转载于:https://www.cnblogs.com/superxuezhazha/p/5693036.html