博客
关于我
强烈建议你试试无所不能的chatGPT,快点击我
POJ-3069
阅读量:5051 次
发布时间:2019-06-12

本文共 2341 字,大约阅读时间需要 7 分钟。

Saruman's Army
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 10994   Accepted: 5555

Description

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, …, xn of 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

0 310 20 2010 770 30 1 7 15 20 50-1 -1

Sample Output

24

Hint

In the first test case, Saruman may place a palantir at positions 10 and 20. Here, note that a single palantir with range 0 can cover both of the troops at position 20.

In the second test case, Saruman can place palantirs at position 7 (covering troops at 1, 7, and 15), position 20 (covering positions 20 and 30), position 50, and position 70. Here, note that palantirs must be distributed among troops and are not allowed to “free float.” Thus, Saruman cannot place a palantir at position 60 to cover the troops at positions 50 and 70.

 

题意:

给n个点,在每个点的r范围内必须要有一个标记。

 

贪心,从最左边开始找r内最远的点。

 

AC代码:

1 //#include
2 #include
3 #include
4 #include
5 using namespace std; 6 7 int a[1010]; 8 9 int main(){10 ios::sync_with_stdio(false);11 int r,n;12 while(cin>>r>>n&&r>=0&&n>=0){13 memset(a,0,sizeof(a));14 for(int i=0;i
>a[i];16 }17 sort(a,a+n);18 int temp=0,ans=0;19 while(temp

 

转载于:https://www.cnblogs.com/Kiven5197/p/7265622.html

你可能感兴趣的文章
第九次团队作业-测试报告与用户使用手册
查看>>
Equal Sides Of An Array
查看>>
CentOS笔记-用户和用户组管理
查看>>
Mongodb 基本命令
查看>>
Qt中QTableView中加入Check列实现
查看>>
“富豪相亲大会”究竟迷失了什么?
查看>>
控制文件的备份与恢复
查看>>
返回代码hdu 2054 A==B?
查看>>
Flink独立集群1
查看>>
iOS 8 地图
查看>>
20165235 第八周课下补做
查看>>
[leetcode] 1. Two Sum
查看>>
iOS 日常工作之常用宏定义大全
查看>>
PHP的SQL注入技术实现以及预防措施
查看>>
MVC Razor
查看>>
软件目录结构规范
查看>>
Windbg调试Sql Server 进程
查看>>
linux调度器系列
查看>>
mysqladmin
查看>>
解决 No Entity Framework provider found for the ADO.NET provider
查看>>