Codeforces Round #256 (Div. 2/C)/Codeforces448C_Painting Fence(分治)

2024-09-04 08:38

本文主要是介绍Codeforces Round #256 (Div. 2/C)/Codeforces448C_Painting Fence(分治),希望对大家解决编程问题提供一定的参考价值,需要的开发者们随着小编来一起学习吧!

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,,,刷子可以在横着和竖着刷,不能跳着刷,,,

如果是竖着刷,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.



这篇关于Codeforces Round #256 (Div. 2/C)/Codeforces448C_Painting Fence(分治)的文章就介绍到这儿,希望我们推荐的文章对编程师们有所帮助!



http://www.chinasem.cn/article/1135534

相关文章

Codeforces Round #240 (Div. 2) E分治算法探究1

Codeforces Round #240 (Div. 2) E  http://codeforces.com/contest/415/problem/E 2^n个数,每次操作将其分成2^q份,对于每一份内部的数进行翻转(逆序),每次操作完后输出操作后新序列的逆序对数。 图一:  划分子问题。 图二: 分而治之,=>  合并 。 图三: 回溯:

Codeforces Round #261 (Div. 2)小记

A  XX注意最后输出满足条件,我也不知道为什么写的这么长。 #define X first#define Y secondvector<pair<int , int> > a ;int can(pair<int , int> c){return -1000 <= c.X && c.X <= 1000&& -1000 <= c.Y && c.Y <= 1000 ;}int m

Codeforces Beta Round #47 C凸包 (最终写法)

题意慢慢看。 typedef long long LL ;int cmp(double x){if(fabs(x) < 1e-8) return 0 ;return x > 0 ? 1 : -1 ;}struct point{double x , y ;point(){}point(double _x , double _y):x(_x) , y(_y){}point op

Codeforces Round #113 (Div. 2) B 判断多边形是否在凸包内

题目点击打开链接 凸多边形A, 多边形B, 判断B是否严格在A内。  注意AB有重点 。  将A,B上的点合在一起求凸包,如果凸包上的点是B的某个点,则B肯定不在A内。 或者说B上的某点在凸包的边上则也说明B不严格在A里面。 这个处理有个巧妙的方法,只需在求凸包的时候, <=  改成< 也就是说凸包一条边上的所有点都重复点都记录在凸包里面了。 另外不能去重点。 int

Codeforces 482B 线段树

求是否存在这样的n个数; m次操作,每次操作就是三个数 l ,r,val          a[l] & a[l+1] &......&a[r] = val 就是区间l---r上的与的值为val 。 也就是意味着区间[L , R] 每个数要执行 | val 操作  最后判断  a[l] & a[l+1] &......&a[r] 是否= val import ja

CSS实现DIV三角形

本文内容收集来自网络 #triangle-up {width: 0;height: 0;border-left: 50px solid transparent;border-right: 50px solid transparent;border-bottom: 100px solid red;} #triangle-down {width: 0;height: 0;bor

创建一个大的DIV,里面的包含两个DIV是可以自由移动

创建一个大的DIV,里面的包含两个DIV是可以自由移动 <body>         <div style="position: relative; background:#DDF8CF;line-height: 50px"> <div style="text-align: center; width: 100%;padding-top: 0px;"><h3>定&nbsp;位&nbsp;

Codeforces Round 971 (Div. 4) (A~G1)

A、B题太简单,不做解释 C 对于 x y 两个方向,每一个方向至少需要 x / k 向上取整的步数,取最大值。 由于 x 方向先移动,假如 x 方向需要的步数多于 y 方向的步数,那么最后 y 方向的那一步就不需要了,答案减 1 代码 #include <iostream>#include <algorithm>#include <vector>#include <string>

CF#271 (Div. 2) D.(dp)

D. Flowers time limit per test 1.5 seconds memory limit per test 256 megabytes input standard input output standard output 题目链接: http://codeforces.com/contest/474/problem/D We s