UVALive 6908 Electric Bike dp
Electric Bike
题目连接:
Description
Two years ago, Putri bought an electric bike (e-bike). She likes e-bike a lot since it can assist her in
cycling through steep roads when commuting to work. Over time, the battery capacity decreases. Now,
her e-bike battery capacity is so low that she needs to carefully plan when to turn on the assist and at
which level; different level of assist consumes different amount of energy and produces different assist
power.
Putri divides the road that she travels from her home to her workplace into N segments where each
segment has a steepness level. For example, the figure below shows a road with N = 7 segments.
Segment #1 #2 #3 #4 #5 #6 #7
Steepness 10 3 0 1 2 15 9
From her home, Putri has to travel from the first road segment, to the second road segment, and so on,
until the last segment to reach her work place. In the example above, the first segment has steepness of
10 which means Putri has to pedal her bike with 10 unit of energy. Her e-bike has fixed 4 assist levels
where each level generates different power, as shown in the following table:
Assist Level 0 1 2 3
Assist Power 0 4 8 11
Assist level L will consume L unit of energy from the battery and will give Putri additional pedaling
assist power according to the table above. Putri can only change the assist level to level L if the battery
has at least L energy left and she is at the beginning of a road segment. Setting an assist level where
the generated power is higher than the road steepness will cause the excess energy power to be wasted.
For example, if Putri sets the assist level L = 2 at the beginning of the first road segment (with
steepness 10), then she only needs to pedal her bike with 2 unit of energy instead of 10 (since her ebike
is assisting her with 8 unit of energy) to advance to the second road segment. If Putri sets the assist
level L = 3, her e-bike generates more power to handle the steepness 10, thus she does not need to
pedal at all, and the excess energy is wasted.
Putri can change the assist level instantly before entering a road segment, however, she does not
want to change the assist level more than K times (it’s too tiring). If there is not enough energy in
the battery to support the selected assist level for the road segment, the e-bike will shutdown at the
beginning of the road segment and Putri has to pedal through the rest of the road segments, i.e. the
assist level automatically set to 0 for the rest of the journey. Note that Putri can change her e-bike
assist level (given it’s still less than K) at the beginning of the road segment to avoid shutdown by
force. Initially at her home, the assist level is set to 0 and the battery is fully charged with E unit of
energy. Putri wants to know the minimum energy she will need to pedal the bike to reach the workplace
if she utilizes her e-bike optimally.
Input
The first line of input contains T (T ≤ 100) denoting the number of cases. Each case begins with three
integers: N, K, and E in a line (1 ≤ N ≤ 1, 000; 0 ≤ K ≤ 10; 0 ≤ E ≤ 50) as described in the problem
statement above. The next line contains N non-negative integers denoting the steepness level of i-th
segment where i = 1 . . . N respectively. The steepness level of any road segment is at most 15.
Output
For each case, output ‘Case #X: Y ’, where X is the case number starts from 1 and Y is the minimum
energy Putri needs to pedal the e-bike from her home to her workplace.
Explanation for 1st sample case:
Putri changes the assist level to 1 at (the beginning of) road segment #2, then change the assist
level to 3 at road segment #6. Thus, she needs to pedal with 10 unit of energy for road segment #1 and
4 unit of energy for road segment #6. Note that if she changes the assist level to 1 at road segment #1
and then to assist level 3 at road segment #6, then at the beginning of road segment #7 the battery
only has 2 unit of energy left and will automatically shutdown to assist level 0, thus Putri has to pedal
with 9 energy for road segment #7.
Explanation for 2nd sample case:
Putri changes the assist level to 3 at road segment #1 and then changes to assist level 2 at road
segment #2. Thus, she only needs to pedal 7 unit of energy for road segment #6 and 1 unit of energy
for road segment #7.
Explanation for 3rd sample case:
Putri changes the assist level to 3 at road segment #1, then changes to assist level 1 at road segment
2, finally changes to assist level 3 at road segment #6. Thus, she only needs to pedal 4 unit of energy
for road segment #6.
Sample Input
5
7 2 10
10 3 0 1 2 15 9
7 2 15
10 3 0 1 2 15 9
7 3 15
10 3 0 1 2 15 9
5 2 5
11 15 1 14 12
15 8 30
2 14 6 1 2 13 14 12 13 12 7 12 1 2 10
Sample Output
Case #1: 14
Case #2: 8
Case #3: 4
Case #4: 34
Case #5: 18
Hint
题意
有n个线段,每个线段长a[i]米,然后你骑着一个电瓶车,电瓶车只有E的电,然后一档需要1能量,可以帮你爬4米,2档需要2能量,可以帮你爬8米,3档需要3能量,可以帮你爬11米。
然后你不够的时候,就只能自己骑上去。
你最多修改k次档位。
如果你能量不够的话,那么你就强行退回了0档,且不能变成其他档。
题解:
虽然是个DP,但实际上是一个模拟题,特别烦……
你就把题目中所有的变量,都当成dp状态跑一边就行了。……
其实可能写成记忆化搜索的形式,更好一点。
代码
#include<bits/stdc++.h>
using namespace std;
int dp[1005][11][51][4][2];
int p[4]={0,4,8,11};
int a[1005];
int cas;
//第i个位置,j次换档,花了k,当前档为t
void solve(){
int n,k,e;
scanf("%d%d%d",&n,&k,&e);
for(int i=1;i<=n;i++)
scanf("%d",&a[i]);
for(int i=0;i<1005;i++)
for(int j=0;j<11;j++)
for(int k=0;k<51;k++)
for(int t=0;t<4;t++)
for(int i2=0;i2<2;i2++)
dp[i][j][k][t][i2]=1e9;
dp[0][0][0][0][0]=0;
dp[0][0][0][0][1]=0;
for(int i=0;i<n;i++){
for(int j=0;j<=k;j++){
for(int K=0;K<=e;K++){
for(int t=0;t<4;t++){
for(int m=0;m<4;m++){
dp[i+1][j][K][0][1]=min(dp[i+1][j][K][0][1],dp[i][j][K][t][0]+a[i+1]);
dp[i+1][j][K][0][1]=min(dp[i+1][j][K][0][1],dp[i][j][K][0][1]+a[i+1]);
if(t==m){
if(K+m>e)
dp[i+1][j][K][0][1]=min(dp[i+1][j][K][0][1],dp[i][j][K][t][0]+a[i+1]);
else
dp[i+1][j][K+m][m][0]=min(dp[i+1][j][K+m][m][0],dp[i][j][K][t][0]+max(0,a[i+1]-p[m]));
}else{
if(j==k){
continue;
}else if(K+m>e)
dp[i+1][j+1][K][0][1]=min(dp[i+1][j+1][K][0][1],dp[i][j][K][t][0]+a[i+1]);
else
dp[i+1][j+1][K+m][m][0]=min(dp[i+1][j+1][K+m][m][0],dp[i][j][K][t][0]+max(0,a[i+1]-p[m]));
}
}
}
}
}
}
int Ans = 1000000000;
for(int j=0;j<=k;j++)
for(int K=0;K<=e;K++)
for(int t=0;t<4;t++)
for(int i2=0;i2<2;i2++)
Ans=min(Ans,dp[n][j][K][t][i2]);
printf("Case #%d: %d\n",++cas,Ans);
}
int main(){
int t;
scanf("%d",&t);
while(t--)solve();
return 0;
}
UVALive 6908 Electric Bike dp的更多相关文章
- UVALive 6908---Electric Bike(DP或记录型深搜)
题目链接 https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_ ...
- UVALive - 6952 Cent Savings dp
题目链接: http://acm.hust.edu.cn/vjudge/problem/116998 Cent Savings Time Limit: 3000MS 问题描述 To host a re ...
- UVALive - 6529 找规律+dp
题目链接: http://acm.hust.edu.cn/vjudge/problem/47664 Eleven Time Limit: 5000MS 问题描述 In this problem, we ...
- UVaLive 6801 Sequence (计数DP)
题意:给定一个序列,有 n 个数,只有01,然后你进行k次操作,把所有的1变成0,求有多种方法. 析:DP是很明显的,dp[i][j] 表示进行第 i 次操作,剩下 j 个1,然后操作就两种,把1变成 ...
- UVaLive 6697 Homework Evaluation (DP)
题意:给出一个长字符串,再给一个短字符串,进行匹配,如果第i个恰好匹配,则 +8,:如果不匹配,可以给长或短字符串添加-,先后匹配,这样-3, 连续的长字符串添加-,需要减去一个4:也可不给添加-,则 ...
- UVaLive 7374 Racing Gems (DP,LIS)
题意:以辆赛车可以从x轴上任意点出发,他的水平速度允许他向每向上移动v个单位,就能向左或向右移动v/r个单位(也就是它的辐射范围是个等腰三角形) 现在赛车从x轴出发,问它在到达终点前能吃到的最多钻石. ...
- UVALive 6947 Improvements(DP+树状数组)
[题目链接] https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=sho ...
- UVaLive 3490 Generator (KMP + DP + Gauss)
题意:随机字母组成一个串,有一个目标串,当这个由随机字母组成的串出现目标串就停止,求这个随机字母组成串的期望长度. 析:由于只要包含目标串就可以停止,所以可以先把这个串进行处理,也就是KMP,然后dp ...
- UVALive 5983 二分答案+dp
想了很久都想不出怎么dp,然后发现有些例子,如果你开始不确定起始值的话,是不能dp的,每种状态都有可能,所以只能二分一个答案,确定开始的val值,来dp了. #include <cstdio&g ...
随机推荐
- ZCMU 1894: Power Eggs
http://acm.zcmu.edu.cn/JudgeOnline/problem.php?id=1894 题意: 有M个鹰蛋,N层楼,鹰蛋的硬度是E,也就是说在1~E层楼扔下去不会碎,E+1层楼扔 ...
- bzoj千题计划203:bzoj3994: [SDOI2015]约数个数和
http://www.lydsy.com/JudgeOnline/problem.php?id=3994 设d(x)为x的约数个数,给定N.M,求 用到的一个结论: 证明: 枚举n的约数i,枚举m的约 ...
- CSS规范 - 优化方案--(来自网易)
值缩写 缩写值可以减少CSS文件大小,并增加可读性和可维护性. 但并非所有的值都必须缩写,因为当一个属性的值缩写时,总是会将所有项都设置一遍,而有时候我们不希望设置值里的某些项. /* 比如我们用下面 ...
- CSS3 - chrome,傲游,360极速浏览器不支持小于12px的字号的解决办法
google流量器chrome,傲游,360极速浏览器都是基于webkit内核浏览器,默认不支持小于font-size小于12px 的字号,即定义font-size小于12px时会发现字体大小依然是1 ...
- 记关于webpack4下css提取打包去重复的那些事
注意使用vue-cli3(webpack4),默认小于30k不会抽取为公共文件,包括css和js,已测试 经过2天的填坑,现在终于有点成果 环境webpack4.6 + html-webpack-pl ...
- python 入门基础4 --数据类型及内置方法
今日目录: 零.解压赋值+for循环 一. 可变/不可变和有序/无序 二.基本数据类型及内置方法 1.整型 int 2.浮点型float 3.字符串类型 4.列表类型 三.后期补充内容 零.解压赋值+ ...
- oracle 建用户
create user username identified by password; grant dba to username; 注意当对用户赋予resource角色时将同时赋予unlimite ...
- Project Euler Problem3
Largest prime factor Problem 3 The prime factors of 13195 are 5, 7, 13 and 29. What is the largest p ...
- 004_on-my-zsh漂亮的shell
一. http://www.cnblogs.com/GarveyCalvin/p/4301235.html 二. 前言:Zsh可配置性强,用户可以自定义配置,个性化强.Zsh tab补全更强大,该功能 ...
- Hitcon 2016 Pwn赛题学习
PS:这是我很久以前写的,大概是去年刚结束Hitcon2016时写的.写完之后就丢在硬盘里没管了,最近翻出来才想起来写过这个,索性发出来 0x0 前言 Hitcon个人感觉是高质量的比赛,相比国内的C ...