一、热身 [Cloned] U - Long Distance Racing

最新推荐文章于 2018-08-10 09:12:08 发布

npuyan

最新推荐文章于 2018-08-10 09:12:08 发布

阅读量212

点赞数

CC 4.0 BY-SA版权

分类专栏：暑假练习

本文链接：https://blog.youkuaiyun.com/npuyan/article/details/81347151

暑假练习专栏收录该内容

60 篇文章

订阅专栏

本文介绍了一个跑步训练场景下的路径规划问题。目标是在限定时间内完成往返跑步，并尽可能地到达最远距离。文章给出了具体的实现代码，通过计算不同地形的跑步时间来确定最大可行距离。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

原题：

Bessie is training for her next race by running on a path that includes hills so that she will be prepared for any terrain. She has planned a straight path and wants to run as far as she can -- but she must be back to the farm within Mseconds (1 ≤ M ≤ 10,000,000).

The entire path she has chosen is T units (1 ≤ T ≤ 100,000) in length and consists of equal-length portions that are uphill, flat, or downhill. The input data describes path segment i with a single character Si that is u, f, or d, indicating respectively uphill, flat, or downhill.

Bessie takes U seconds (1 ≤ U ≤ 100) to run one unit of uphill path, F (1 ≤ F≤ 100) seconds for a unit of flat path, and D (1 ≤ D ≤ 100) seconds for a unit of downhill path. Note that, when returning home, uphill paths become downhill paths and downhill paths become uphill paths.

Find the farthest distance Bessie can get from the farm and still make it back in time.

题意：

一个人跑步，要在规定的时间T内跑一个来回，然后给出一段路，问她最多能跑到哪里才能在规定时间内回来。给出的路段有平路，上坡，下坡，三种地形跑步的速度不同，因为是来回，所以同一段路跑两次。（超级水）

题解：

依次算路段，平路的话就时间乘2，上坡的话在加一个下坡，下坡的话在加一个上坡，用总时间做限制。

代码：AC

#include<iostream>
using namespace std;
int main()
{
	int M,T,U,F,D;
	cin>>M>>T>>U>>F>>D;
	int time=0;
	int k=0,i;
	char s;
	for(i=0;i<T;i++)
	{
		cin>>s;
		if(time<=M)
		{
			if(s=='u'||s=='d')
				time+=U+D;
			else if(s=='f')
				time+=2*F;
			k++;
		}
	}
	cout<<k-1<<endl;
	return 0;
}