Cow Cycling 动态规划
1552: Cow Cycling
时间限制(普通/Java):1000MS/10000MS 内存限制:65536KByte
总提交: 39 测试通过:20
描述
The cow bicycling team consists of N (1 <= N <= 20) cyclists. They wish to determine a race strategy which will get one of them across the finish line as fast as possible.
Like everyone else, cows race bicycles in packs because that's the most efficient way to beat the wind. While travelling at x laps/minute (x is always an integer), the head of the pack expends x*x energy/minute while the rest of pack drafts behind him using only x energy/minute. Switching leaders requires no time though can only happen after an integer number of minutes. Of course, cows can drop out of the race at any time.
The cows have entered a race D (1 <= D <= 100) laps long. Each cow has the same initial energy, E (1 <= E <= 100).
What is the fastest possible finishing time? Only one cow has to cross the line. The finish time is an integer. Overshooting the line during some minute is no different than barely reaching it at the beginning of the next minute (though the cow must have the energy left to cycle the entire minute). N, D, and E are integers.
输入
A single line with three integers: N, E, and D
输出
A single line with the integer that is the fastest possible finishing time for the fastest possible cow. Output 0 if the cows are not strong enough to finish the race.
样例输入
3 30 20
样例输出
7
提示
as shown in this chart:
leader E
pack total used this
time leader speed dist minute
1 1 5 5 25
2 1 2 7 4
3 2* 4 11 16
4 2 2 13 4
5 3* 3 16 9
6 3 2 18 4
7 3 2 20 4
* = leader switch
题意:有N头奶牛,每头奶牛的能量是E,现在有一个任务是跑完D圈,但是只要有一头奶牛完成这个任务就算通过。每次需要有一头奶牛领跑,其他的奶牛可以选择继续跟着跑或者离开队伍。领跑的奶牛能量
消耗是x*x laps/min,跟跑的能量消耗是x laps/min,然后让你计算最短需要多少时间完成任务。
题解:有N头奶牛,那么当N-1头奶牛都领跑过,那么最后一头奶牛去完成任务就成了。
状态:wxl[i][j][t],第i头奶牛跑 j 圈,消耗t能量所花费的最短时间。
状态转移方程:wxl[i+1][j][j]=min(wxl[i+1][j][j],wxl[i][j][t]);//换奶牛领跑不消耗时间 wxl[i][j+l][l*l+t]=min(wxl[i][j][t]+1,wxl[i][j+l][l*l+t]);
#include "bits/stdc++.h"
using namespace std;
#define INF 0x3f3f3f3f
int wxl[][][];//存状态,第i头奶牛跑j圈,第i头奶牛消耗t所花费的最小时间
int main()
{
ios::sync_with_stdio(false);
cin.tie();cout.tie();//输入输出加速
int n,e,d,i,j,t,l,sum=INF;
cin>>n>>e>>d;
for(i=;i<=n;++i)for(j=;j<=d;++j)for(t=;t<=e;++t)wxl[i][j][t]=INF;
wxl[][][]=;
for(i=;i<=n;++i)for(j=;j<=d;++j)for(t=;t<=e;++t)
{
if(wxl[i][j][t]==INF)continue;
for(l=;l+j<=d&&l*l+t<=e;++l)wxl[i][j+l][l*l+t]=min(wxl[i][j][t]+,wxl[i][j+l][l*l+t]);
wxl[i+][j][j]=min(wxl[i+][j][j],wxl[i][j][t]);//换奶牛领跑不消耗时间
}
for(i=;i<=e;++i)sum=min(sum,wxl[n][d][i]);
cout<<sum<<endl;//当完不成任务时输出wxl[n][d][0];
}
//状态转移方程是wxl[i][t+l][l*l+t]=min(wxl[i][j][t]+1,wxl[i][t+l][l*l+t]);wxl[i+1][j][j]=min(wxl[i+1][j][j],wxl[i][j][t]);
Cow Cycling 动态规划的更多相关文章
- [USACO2002][poj1946]Cow Cycling(dp)
Cow CyclingTime Limit: 1000MS Memory Limit: 30000KTotal Submissions: 2468 Accepted: 1378Description ...
- POJ 1946 Cow Cycling
Cow Cycling Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 2516 Accepted: 1396 Descr ...
- POJ3267——The Cow Lexicon(动态规划)
The Cow Lexicon DescriptionFew know that the cows have their own dictionary with W (1 ≤ W ≤ 600) wor ...
- POJ3176——Cow Bowling(动态规划)
Cow Bowling DescriptionThe cows don't use actual bowling balls when they go bowling. They each take ...
- POJ - 3176 Cow Bowling 动态规划
动态规划:多阶段决策问题,每步求解的问题是后面阶段问题求解的子问题,每步决策将依赖于以前步骤的决策结果.(可以用于组合优化问题) 优化原则:一个最优决策序列的任何子序列本身一定是相当于子序列初始和结束 ...
- POJ 1946 Cow Cycling(抽象背包, 多阶段DP)
Description The cow bicycling team consists of N (1 <= N <= 20) cyclists. They wish to determi ...
- 【BZOJ3939】[Usaco2015 Feb]Cow Hopscotch 动态规划+线段树
[BZOJ3939][Usaco2015 Feb]Cow Hopscotch Description Just like humans enjoy playing the game of Hopsco ...
- PKU 3267 The Cow Lexicon(动态规划)
题目大意:给定一个字符串和一本字典,问至少需要删除多少个字符才能匹配到字典中的单词序列.PS:是单词序列,而不是一个单词 思路: ...
- poj 3267 The Cow Lexicon (动态规划)
The Cow Lexicon Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 8167 Accepted: 3845 D ...
随机推荐
- Modelsimobjects空白无显示问题和ISim仿真时出现空白
困扰朕长达一周的问题得以解决!!!!! 发生这种情况的根源是win10自带的防火墙的问题.只有关闭防火墙,再重新打开软件进行仿真就能出现正常的仿真界面. 关闭防火墙的方法为:控制面板>>系 ...
- pwn学习日记Day2 基础知识积累
知识杂项 shell-storm.org的shellcode数据库 使用pwntools库把shellcode作为输入传递给程序,尝试使用io.interactive()与程序进行交互,发现可以执行s ...
- javascript中计算两个时间日期间隔的天数
<script> /* 计算两个日期的时间间隔天数 */ //时间字符串的格 ...
- spring boot中使用@Async实现异步调用任务
本篇文章主要介绍了spring boot中使用@Async实现异步调用任务,小编觉得挺不错的,现在分享给大家,也给大家做个参考.一起跟随小编过来看看吧 什么是“异步调用”? “异步调用”对应的是“同步 ...
- The type com.google.protobuf.GeneratedMessageV3$Builder cannot be resolved. It is indirectly referenced from required .classfiles
在做项目的时候,导入了几个类,导入了相关的jar. 结果在package处报了The type com.google.protobuf.GeneratedMessageV3$Builder canno ...
- 读HashMap 源码(jdk11)的见解
如果想系统详细的了解HashMap请移步各大佬博客.这篇文章只是个人的一些见解. 数组+链表 或 数组+红黑树.这种说法感觉有迷惑性. 之前看博客都说 HashMap 的存储是数组+链表(jdk6), ...
- 解决Linux 环境 GLIBCXX_3.4.15' not found问题
升级Centos系统之后,运行filezilla时,出现如下错误的提示信息: ./filezilla: /usr/lib/libstdc++.so.6: version `GLIBCXX_3.4.15 ...
- Fetch和ajax的比较和区别
传统 Ajax 已死,Fetch 永生 Ajax 不会死,传统 Ajax 指的是 XMLHttpRequest(XHR),未来现在已被 Fetch 替代. 最近把阿里一个千万级 PV 的数据产品全 ...
- 如何在linux上查看tomcat的端口号
1.先到tomcat配置文件查看tomcat的端口是什么,配置文件一般是:$CATALINA_HOME/conf/server这个文件,查找<Connector port="8080& ...
- Linux基础命令---free显示内存使用
free free指令用来显示内存的使用情况,显示系统中可用和已使用的物理和交换内存的总量,以及内核使用的缓冲区.应该忽略共享内存列:它已经过时了. 此命令的适用范围:RedHat.RHEL.Ubun ...