POJ3069 Saruman's Army【贪心】
Saruman the White must lead his army along a straight path from Isengard to Helm’s Deep. To keep track of his forces, Saruman distributes seeing stones, known as palantirs, among the troops. Each palantir has a maximum effective range of R units, and must be carried by some troop in the army (i.e., palantirs are not allowed to “free float” in mid-air). Help Saruman take control of Middle Earth by determining the minimum number of palantirs needed for Saruman to ensure that each of his minions is within R units of some palantir.
Input
The input test file will contain multiple cases. Each test case begins with a single line containing an integer R, the maximum effective range of all palantirs (where 0 ≤ R ≤ 1000), and an integer n, the number of troops in Saruman’s army (where 1 ≤ n ≤ 1000). The next line contains n integers, indicating the positions x1, …, xnof each troop (where 0 ≤ xi ≤ 1000). The end-of-file is marked by a test case with R = n = −1.
Output
For each test case, print a single integer indicating the minimum number of palantirs needed.
Sample Input
0 3
10 20 20
10 7
70 30 1 7 15 20 50
-1 -1
Sample Output
2
4
思路:从最左边开始,找r范围内最右边的点,然后在超出r范围的最左边点当做起始点继续寻找。
#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=1005;
int main()
{
int n,r, arr[maxn];
while(scanf("%d%d",&r,&n)!=EOF && n+r!=-2)
{
for(int i=0;i<n;++i)
scanf("%d",&arr[i]);
sort(arr,arr+n);
int i=0,ans=0;
while(i<n)
{
int s=arr[i++];
while(i<n && arr[i]<=s+r)
++i;
int p=arr[i-1];
while(i<n && arr[i]<=p+r)
++i;
++ans;
}
printf("%d\n",ans);
}
return 0;
}
POJ3069 Saruman's Army【贪心】的更多相关文章
- POJ 3617 Best Cow Line ||POJ 3069 Saruman's Army贪心
带来两题贪心算法的题. 1.给定长度为N的字符串S,要构造一个长度为N的字符串T.起初,T是一个空串,随后反复进行下面两个操作:1.从S的头部删除一个字符,加到T的尾部.2.从S的尾部删除一个字符,加 ...
- poj3069 Saruman's Army
http://poj.org/problem?id=3069 Saruman the White must lead his army along a straight path from Iseng ...
- POJ 3069 Saruman's Army(贪心)
Saruman's Army Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u Sub ...
- poj 3069 Saruman's Army 贪心模拟
Saruman's Army Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 18794 Accepted: 9222 D ...
- B - B Saruman's Army(贪心)
在一条直线上,有n个点.从这n个点中选择若干个,给他们加上标记.对于每一个点,其距离为R以内的区域里必须有一个被标记的点.问至少要有多少点被加上标记 Saruman the White must le ...
- poj3069 Saruman's Army(贪心)
https://vjudge.net/problem/POJ-3069 弄清楚一点,第一个stone的位置,考虑左右两边都要覆盖R,所以一般情况下不会在左边第一个(除非前两个相距>R). 一开始 ...
- poj 3069 Saruman's Army 贪心 题解《挑战程序设计竞赛》
地址 http://poj.org/problem?id=3069 题解 题目可以考虑贪心 尽可能的根据题意选择靠右边的点 注意 开始无标记点 寻找左侧第一个没覆盖的点 再来推算既可能靠右的标记点为一 ...
- Saruman's Army(贪心)
Saruman the White must lead his army along a straight path from Isengard to Helm’s Deep. To keep tra ...
- Saruman's Army(POJ3069)
Description Saruman the White must lead his army along a straight path from Isengard to Helm’s Deep. ...
随机推荐
- jsp中EL表达式不起作用的问题
jsp中EL表达式不起作用的问题 进行springmvc的@ExceptioinHandler调试,竟然是el表达式的问题, 学习了:http://blog.csdn.net/wolf_soul/ar ...
- 重学C++ (十一) OOP面向对象编程(2)
转换与继承 本节主要须要区分的是: 基类和派生类的转换: 引用(指针)的转换和对象的转换. 1.每一个派生类对象包括一个基类部分.因此.能够像使用基类对象一样在派生类对象上执行操作. 基于这一点,能够 ...
- 【Linux】Linux下配置apache
一.获取软件: http://httpd.apache.org/ httpd-2.4.10.tar.gz 二.安装步骤: 解压源文件: 1) tar zvxf httpd-2.4.10.tar. ...
- SpringMVC + hibernate 配置文件
web.xml <?xml version="1.0" encoding="UTF-8"?> <web-app xmlns="htt ...
- string[][]和string[,] 以及 int[][]和int[,]
string[][]和string[,] http://www.codewars.com/kata/56f3a1e899b386da78000732/train/csharp Write a func ...
- C# 操作 INI 自己工作笔记(对文本框的操作)
using System;using System.Collections.Generic;using System.ComponentModel;using System.Data;using Sy ...
- 什么是 less? 如何使用 less?
什么是 Less? Less 是一门 CSS 预处理语言,它扩充了 CSS 语言,增加了诸如变量.混合(mixin).嵌套.函数等功能,让 CSS 更易编写.维护等. 本质上,Less 包含一套自定义 ...
- PCB 工程系统 模拟windows域帐号登入
一.需求描述: 对于PCB制造企业来说,基本都采用建立共享目享+域名管控权限,好像别的大多数行业都是这样的吧.呵呵 在实际应用中,经常会有这样的问题,自己登入的帐号没有共享目录的权限,但又想通过程序实 ...
- [Apple开发者帐户帮助]三、创建证书(7)创建证书签名请求
Mac上的Keychain Access允许您创建证书签名请求(CSR). 启动位于的Keychain Access /Applications/Utilities. 选择Keychain Acces ...
- Django day14(一) cookie
一: Cookie 1. Cookie是什么?存储在客户端浏览器上的键值对 2. 原理: 是服务器产生,发给客户端浏览器,浏览器保存起来,下次发请求,会携带这个键值对到服务器 4. Cookie的覆 ...