zoukankan      html  css  js  c++  java
  • Painting Fence

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

    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 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 Input

    Input
    5
    2 2 1 2 1
    Output
    3
    Input
    2
    2 2
    Output
    2
    Input
    1
    5
    Output
    1


    #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
    ", 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。

  • 相关阅读:
    MySQL视图更新
    JavaScript经典作用域问题
    进程间通信的几种方式
    Vue(MVVM)、React(MVVM)、Angular(MVC)对比
    CDN(Content Delivery Network)技术原理概要
    单点登录实现原理(SSO)
    composer 实现自动加载原理
    PHP 反射的简单使用
    php7安装php-redis扩展
    Git 简单入门(二)
  • 原文地址:https://www.cnblogs.com/superxuezhazha/p/5693036.html
Copyright © 2011-2022 走看看