解题报告

给篱笆上色,要求步骤最少,篱笆怎么上色应该懂吧,。,刷子能够在横着和竖着刷,不能跳着刷,,,

假设是竖着刷,应当是篱笆的条数,横着刷的话。就是刷完最短木板的长度,再接着考虑没有刷的木板,,。

递归调用,,。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define inf 999999999999999
using namespace std;
long long n,num[5010],tt;
void dfs(long long s,long long t)
{
long long ma=0,mi=inf;
int i,j;
for(i=s; i<=t; i++)
{
if(ma<num[i])
ma=num[i];
if(mi>num[i])
mi=num[i];
}
if(mi==ma)
{
tt+=min(mi,t-s+1);
return ;
}
for(i=s; i<=t; i++)
num[i]-=mi;
tt+=min(mi,t-s);
for(i=s; i<=t; i++)
{
if(num[i]>0)
{
for(j=i; j<=t; j++)
{
if(num[j]==0||(j==t&&num[j]>0))
{
long long kk=tt;
if(j==t&&num[j]>0)
{
dfs(i,j);
if(tt-kk>(j-i+1))
tt=kk+(j-i+1);
}
else
{
dfs(i,j-1);
if(tt-kk>(j-i))
tt=kk+(j-i);
}
i=j;
break;
}
}
}
}
}
int main()
{
int i,j;
scanf("%lld",&n);
for(i=1; i<=n; i++)
scanf("%lld",&num[i]);
dfs(1,n);
printf("%lld\n",min(tt,n));
return 0;
}

Painting Fence
time limit per test

1 second

memory limit per test

512 megabytes

input

standard input

output

standard output

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.


Codeforces Round #256 (Div. 2/C)/Codeforces448C_Painting Fence(分治)的更多相关文章

  1. Codeforces Round #256 (Div. 2) C. Painting Fence(分治贪心)

    题目链接:http://codeforces.com/problemset/problem/448/C C. Painting Fence time limit per test 1 second m ...

  2. Codeforces Round #256 (Div. 2) C. Painting Fence 或搜索DP

    C. Painting Fence time limit per test 1 second memory limit per test 512 megabytes input standard in ...

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

    C. Painting Fence Bizon the Champion isn't just attentive, he also is very hardworking. Bizon the Ch ...

  4. Codeforces Round #256 (Div. 2) C. Painting Fence (搜索 or DP)

    [题目链接]:click here~~ [题目大意]:题意:你面前有宽度为1,高度给定的连续木板,每次能够刷一横排或一竖列,问你至少须要刷几次. Sample Input Input 5 2 2 1 ...

  5. Codeforces Round #256 (Div. 2) 题解

    Problem A: A. Rewards time limit per test 1 second memory limit per test 256 megabytes input standar ...

  6. Codeforces Round #256 (Div. 2) D. Multiplication Table(二进制搜索)

    转载请注明出处:viewmode=contents" target="_blank">http://blog.csdn.net/u012860063?viewmod ...

  7. Codeforces Round #256 (Div. 2) B. Suffix Structures(模拟)

    题目链接:http://codeforces.com/contest/448/problem/B --------------------------------------------------- ...

  8. Codeforces Round #256 (Div. 2)

    A - Rewards 水题,把a累加,然后向上取整(double)a/5,把b累加,然后向上取整(double)b/10,然后判断a+b是不是大于n即可 #include <iostream& ...

  9. Codeforces Round #256 (Div. 2/B)/Codeforces448B_Suffix Structures(字符串处理)

    解题报告 四种情况相应以下四组数据. 给两字符串,推断第一个字符串是怎么变到第二个字符串. automaton 去掉随意字符后成功转换 array 改变随意两字符后成功转换 再者是两个都有和两个都没有 ...

随机推荐

  1. logstash 使用grok正则解析日志

    http://xiaorui.cc/2015/01/27/logstash%E4%BD%BF%E7%94%A8grok%E6%AD%A3%E5%88%99%E8%A7%A3%E6%9E%90%E6%9 ...

  2. logstash Codec

    Logstash 使用一个名叫FileWatch的Ruby Gem库来监听文件变化,这个库支持glob扩展文件路径, 而且会记录一个叫.sincedb的数据库文件来跟踪被监听日志文件的当前读取位置,所 ...

  3. (十)boost库之多线程

    (十)boost库之多线程 1.创建线程 使用boost库可以方便的创建一个线程,并提供最多支持9个参数的线程函数,相对于void*来说,方便了很多,创建线程主要提供了一下3种方式: 线程库头文件:# ...

  4. 关于App的一些迷思以及一些动画效果开源库的推荐

    http://www.open-open.com/lib/view/open1427856817396.html

  5. RTNETLINK answers: Operation not permitted

    如果出现:RTNETLINK answers: Operation not permitted,那是因为没有权限. 解决办法:su,输入root密码.

  6. #include <hash_set>

    哈希查找,不需要排序,适用于精确查找,比二分查找更快 #define _SILENCE_STDEXT_HASH_DEPRECATION_WARNINGS #include <iostream&g ...

  7. pomelo初探

    最近发现了一个比较好玩的东西pomelo.地址:点击打开链接 这个东西是网易开发的一套基于node.js的高性能,分布式游戏服务器框架.这套框架不仅可以用来开发游戏服务器,也可用于开发高实时web应用 ...

  8. paip.navicat form mysql导入文本文件时CPU占用100%的解决

    paip.navicat form  mysql导入文本文件时CPU占用100%的解决 作者Attilax ,  EMAIL:1466519819@qq.com  来源:attilax的专栏 地址:h ...

  9. IDE idea 更换项目的JDK步骤

    1.如图:

  10. PropertyGrid—添加EventTab

    零.引言 前面一篇文章介绍了如何在PropertyGrid中添加属性Tab,本文主要介绍如何添加事件选项卡.事件在许多对象中都有,尤其是在控件中,如何让对象的事件在PropertyGrid中显示出来呢 ...