poj 3672 Long Distance Racing

Long Distance Racing

Time Limit: 1000MS Memory Limit: 65536KTotal 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 1ufudf

Sample Output

3

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;}