POJ1852
2019-11-11 09:00:29来源:博客园 阅读 ()
POJ1852
题目链接:http://poj.org/problem?id=1852
题目表述:
AntsTime Limit: 1000MS | Memory Limit: 30000K | |
Total Submissions: 33151 | Accepted: 12362 |
Description
An army of ants walk on a horizontal pole of length l cm, each with a constant speed of 1 cm/s. When a walking ant reaches an end of the pole, it immediatelly falls off it. When two ants meet they turn back and start walking in opposite directions. We know the original positions of ants on the pole, unfortunately, we do not know the directions in which the ants are walking. Your task is to compute the earliest and the latest possible times needed for all ants to fall off the pole.Input
The first line of input contains one integer giving the number of cases that follow. The data for each case start with two integer numbers: the length of the pole (in cm) and n, the number of ants residing on the pole. These two numbers are followed by n integers giving the position of each ant on the pole as the distance measured from the left end of the pole, in no particular order. All input integers are not bigger than 1000000 and they are separated by whitespace.Output
For each case of input, output two numbers separated by a single space. The first number is the earliest possible time when all ants fall off the pole (if the directions of their walks are chosen appropriately) and the second number is the latest possible such time.Sample Input
2 10 3 2 6 7 214 7 11 12 7 13 176 23 191
Sample Output
4 8 38 207
Source
Waterloo local 2004.09.19 思路:即求出最大值与最小值即可,最小值就是每个蚂蚁离某一端取最近的,然后在每个最近中选择时间需要最多的那个蚂蚁。注意每个蚂蚁是同时进行运动的。 要求时间最大值时,题目描述说,当两个蚂蚁相遇时,只能各自掉头返回,我们在求最大值时,可以把蚂蚁视作无视相遇,照常穿过,同样也可以求出最大值,即取所需时间最多的那个蚂蚁的时间即可。 代码如下:#include<iostream> #include<cmath> using namespace std; int main() { int n,l,t; cin >> t; while(t--) { int min1 = 0,max1 = 0,x; cin >> l >> n; for(int i = 0;i < n;i++) { cin >> x; if(x <= l / 2) { min1 = max(min1,x); max1 = max(max1,l - x); } else { min1 = max(min1,l - x); max1 = max(max1,x); } } cout << min1 << " " << max1 << endl; } return 0; }
原文链接:https://www.cnblogs.com/biaobiao88/p/11832757.html
如有疑问请与原作者联系
标签:
版权申明:本站文章部分自网络,如有侵权,请联系:west999com@outlook.com
特别注意:本站所有转载文章言论不代表本站观点,本站所提供的摄影照片,插画,设计作品,如需使用,请与原作者联系,版权归原作者所有
- P1358 扑克牌 2020-05-06
- 博弈--巴什博弈 2020-04-24
- Z 字形变换 2020-04-14
- [题记-并查集] 合根植物 - 蓝桥杯 2020-04-07
- 无法正确通过算法题目都是哪些原因造成的? 2020-04-05
IDC资讯: 主机资讯 注册资讯 托管资讯 vps资讯 网站建设
网站运营: 建站经验 策划盈利 搜索优化 网站推广 免费资源
网络编程: Asp.Net编程 Asp编程 Php编程 Xml编程 Access Mssql Mysql 其它
服务器技术: Web服务器 Ftp服务器 Mail服务器 Dns服务器 安全防护
软件技巧: 其它软件 Word Excel Powerpoint Ghost Vista QQ空间 QQ FlashGet 迅雷
网页制作: FrontPages Dreamweaver Javascript css photoshop fireworks Flash