[BZOJ1721][Usaco2006 Mar]Ski Lift 缆车支柱
Description
Farmer Ron in Colorado is building a ski resort for his cows (though budget constraints dictate construction of just one ski lift). The lift will be constructed as a monorail and will connect a concrete support at the starting location to the support at the ending location via some number of intermediate supports, each of height 0 above its land. A straight-line segment of steel connects every pair of adjacent supports. For obvious reasons, each section of straight steel must lie above the ground at all points. Always frugal, FR wants to minimize the number of supports that he must build. He has surveyed the N (2 <= N <= 5,000) equal-sized plots of land the lift will traverse and recorded the integral height H (0 <= H <= 1,000,000,000) of each plot. Safety regulations require FR to build adjacent supports no more than K (1 <= K <= N - 1) units apart. The steel between each pair of supports is rigid and forms a straight line from one support to the next. Help FR compute the smallest number of supports required such that: each segment of steel lies entirely above (or just tangent to) each piece of ground, no two consecutive supports are more than K units apart horizontally, and a support resides both on the first plot of land and on the last plot of land.
科罗拉州的罗恩打算为他的奶牛们建造一个滑雪场,虽然需要的设施仅仅是一部缆车.建造一部缆车,需要从山脚到山顶立若干根柱子,并用钢丝连结它们.你可以认为相对于地面,柱子的高度可以忽略不计.每相邻两根柱子间都有钢丝直接相连.显然,所有钢丝的任何一段都不能在地面之下. 为了节省建造的费用,罗恩希望在工程中修建尽可能少的柱子.他在准备修建缆车的山坡上迭定了N(2≤N≤5000)个两两之间水平距离相等的点,并且测量了每个点的高度H(O≤日≤10^9).并且,按照国家安全标准,相邻两根柱子间的距离不能超过K(1≤K≤N-1)个单位长度.柱子间的钢丝都是笔直的. 罗恩希望你帮他计算一下,在满足下列条件的情况下,他至少要修建多少根柱子:首先,所有的柱子都必须修建在他所选定的点上,且每一段钢丝都必须高于地面或者正好跟地面相切.相邻两根柱子的距离不大于K个单位长度.当然,在第一个点与最后一个点上一定都要修建柱子.
Input
* Line 1: Two space-separate integers, N and K
* Lines 2..N+1: Line i+1 contains a single integer that is the height of plot i.
第1行:两个整数N和K,用空格隔开.
第2到N+1行:每行包括一个正整数,第i+l行的数描述了第i个点的高度.
Output
* Line 1: A single integer equal to the fewest number of lift towers FR needs to build subject to the above constraints
输出一个整数,即罗恩最少需要修建的柱子的数目.
Sample Input
0
1
0
2
4
6
8
6
8
8
9
11
12
Sample Output
#include<iostream>
#include<cstdio>
#include<cstring>
#define inf 1e9
using namespace std;
int n,k;
int h[],f[];
int main()
{
cin>>n>>k;
for(int i=;i<=n;i++) cin>>h[i];
memset(f,0x3f,sizeof(f));
f[]=;
for(int i=;i<=n;i++)
{
double maxn=-inf;
for(int j=i+;j<=min(i+k,n);j++)
{
double now=(double)(h[j]-h[i])/(j-i);
if(now>=maxn)
{
f[j]=min(f[j],f[i]+);
maxn=now;
}
}
}
cout<<f[n];
return ;
}
[BZOJ1721][Usaco2006 Mar]Ski Lift 缆车支柱的更多相关文章
- luogu 4909 [Usaco2006 Mar]Ski Lift 缆车支柱 动态规划
可以出模拟赛T1? #include <bits/stdc++.h> #define N 5002 #define inf 1000000 #define setIO(s) freopen ...
- BZOJ1721 Ski Lift 缆车支柱
Description Farmer Ron in Colorado is building a ski resort for his cows (though budget constraints ...
- 【USACO2006 Mar】滑雪缆车 skilift
[USACO2006 Mar] 滑雪缆车 skilift Time Limit 1000 msMemory Limit 131072 KBytes Description 科罗拉多州的罗恩打算为奶牛建 ...
- [BZOJ1659][Usaco2006 Mar]Lights Out 关灯
[BZOJ1659][Usaco2006 Mar]Lights Out 关灯 试题描述 奶牛们喜欢在黑暗中睡觉.每天晚上,他们的牲口棚有L(3<=L<=50)盏灯,他们想让亮着的灯尽可能的 ...
- Bzoj 1657: [Usaco2006 Mar]Mooo 奶牛的歌声 单调栈
1657: [Usaco2006 Mar]Mooo 奶牛的歌声 Time Limit: 5 Sec Memory Limit: 64 MBSubmit: 631 Solved: 445[Submi ...
- BZOJ1657: [Usaco2006 Mar]Mooo 奶牛的歌声
1657: [Usaco2006 Mar]Mooo 奶牛的歌声 Time Limit: 5 Sec Memory Limit: 64 MBSubmit: 489 Solved: 338[Submi ...
- 1657: [Usaco2006 Mar]Mooo 奶牛的歌声
1657: [Usaco2006 Mar]Mooo 奶牛的歌声 Time Limit: 5 Sec Memory Limit: 64 MBSubmit: 526 Solved: 365[Submi ...
- bzoj1722: [Usaco2006 Mar] Milk Team Select 产奶比赛 树形dp
题目链接 bzoj1722: [Usaco2006 Mar] Milk Team Select 产奶比赛 题解 dp[i][j][0 / 1] 以i为根的子数中 相邻点对选了j个的最大价值 代码 #i ...
- 1722: [Usaco2006 Mar] Milk Team Select 产奶比赛
1722: [Usaco2006 Mar] Milk Team Select 产奶比赛 https://www.lydsy.com/JudgeOnline/problem.php?id=1722 分析 ...
随机推荐
- Configuration注解类 Bean解析顺序
@PropertySource 加载properties @ComponentScan 扫描包 @Import 依赖的class @ImportResource 依赖的xml @Bean 创建bean ...
- tomcat报错-----》Unable to open debugger port IDEA Unable to open debugger port
原因:IDEA配置的端口被占用了 解决方法: 方法一: 查找idea配置的调试端口--查看占用该端口的进程--杀掉进程 方法二:查找idea配置的调试端口--修改调试端口(未被使用的) 基本步骤: 1 ...
- AtCoder Tak and Hotels
题目链接:传送门 题目大意:有 n 个点排成一条直线,每次行动可以移动不超过 L 的距离,每次行动完成必须停在点上, 数据保证有解,有 m 组询问,问从 x 到 y 最少需要几次行动? 题目思路:倍增 ...
- 【BZOJ1189】[HNOI2007]紧急疏散evacuate 动态加边网络流
[BZOJ1189][HNOI2007]紧急疏散evacuate Description 发生了火警,所有人员需要紧急疏散!假设每个房间是一个N M的矩形区域.每个格子如果是'.',那么表示这是一块空 ...
- android studio 运行是,app标题栏不显示
解决办法:让所有的活动都继承 AppCompatActivity就行了,如: public class FirstActivity extends AppCompatActivity{ ... }
- poj 1182 食物链 (带关系的并查集)
食物链 Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 44835 Accepted: 13069 Description 动 ...
- IO流入门-第十章-DataInputStream_DataOutputStream
DataInputStream和DataOutputStream基本用法和方法示例 /* java.io.DataOutputStream 数据字节输出流,带着类型写入 可以将内存中的“int i = ...
- 总结学习! xml与java对象转换 --- JDK自带的JAXB(Java Architecture for XML Binding)
JAXB(Java Architecture for XML Binding) 是一个业界的标准,是一项可以根据XML Schema产生Java类的技术.该过程中,JAXB也提供了将XML实例文档反向 ...
- scrapy spider
spider 定义:在spiders文件夹中由用户自定义,继承scrapy.Spider类或其子类 Spider并没有提供什么特殊的功能. 其仅仅请求给定的 start_urls/start_requ ...
- Django - 回顾(2)- 中介模型
一.中介模型 我们之前学习图书管理系统时,设计了Publish.Book.Author.AuthorDetail这样几张表,其中Book表和Author表是多对多关系,处理类似这样简单的多对多关系时, ...