poj 3672 Long Distance Racing

最新推荐文章于 2019-03-31 20:30:05 发布

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Bessie为了准备她的下一场比赛，在规定的1到10,000,000秒时间内，通过跑步训练跨越不同地形。输入包括最大时间限制、路径长度、上坡、平路和下坡所需时间，输出Bessie能跑的最大距离。

Long Distance Racing

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7326		Accepted: 3907

Description

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 M seconds (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 Sithat 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.

Input

* Line 1: Five space-separated integers: M, T, U, F, and D
* Lines 2..T+1: Line i+1 describes path segment i with a single letter: S_i

Output

* Line 1: A single integer that is the farthest distance (number of units) that Bessie can get from the farm and make it back in time.

Sample Input

13 5 3 2 1
u
f
u
d
f

Sample Output

题意：Bessie要跑步，不过他要在规定的时间m内返回，有三种路况：上坡，平路，下坡；现在有t个路，同时告诉你这三种路况需要的时间u，f，d，问Bessie能走的最远距离；

#include <iostream>
using namespace std;
int main(){
	int m,t,u,f,d;
	char path[100005];
	int distance=0;
	while (cin>>m>>t>>u>>f>>d){
		distance=0;
		for (int i=0;i<t;i++){
			cin>>path[i];
			if (m==0)
				continue;
			if (path[i]=='u'||path[i]=='d'){
				if (m<(u+d))
					m=0;
				else{
					distance++;
					m-=(u+d);
				}
					
			}
			else{
				if (m<f+f)
					m=0;
				else{
					distance++;
					m-=2*f;
				}
			}
		}
		cout<<distance<<endl;	
		
	}
	return 0;
}