Tickets
Description
A good approach, reducing the total time of tickets selling, is let adjacent people buy tickets together. As the restriction of the Ticket Seller Machine, Joe can sell a single ticket or two adjacent tickets at a time.
Since you are the great JESUS, you know exactly how much time needed for every person to buy a single ticket or two tickets for him/her. Could you so kind to tell poor Joe at what time could he go back home as early as possible? If so, I guess Joe would full of appreciation for your help.
Input
1) An integer K(1<=K<=2000) representing the total number of people;
2) K integer numbers(0s<=Si<=25s) representing the time consumed to buy a ticket for each person;
3) (K-1) integer numbers(0s<=Di<=50s) representing the time needed for two adjacent people to buy two tickets together.
Output
Sample Input
2
20 25
40
1
8
Sample Output
08:00:08 am
/*
题意:现在有n个人买票,每个人买票都耗费时间。每个人有两种选择,自己买票,和前面的人或者和后面的人一起买票,给出和前面的人
或者和后面的人一起买票耗费的时间。让你求出最少耗费的时间。 初步思路:dp[i]=max(dp[i],dp[i-1]-use_time[i-1]+double_time[i]);瞎想了一个状态转移方程试着搞一搞 #错误:直觉错了。想法没错,但是方程写错了,dp[i]表示前i个人话费的最少时间,考虑第i个人的时候,和前面的那个人一起买票,还
是自己买票,得出来状态转移方程:dp[i]=max(dp[i-2]+double[i],dp[i-1]+use_time[i-1]) #还是错:正在找原因。。。坑 说了用 am 和 pm 但是数据要求用24进制,另外有一个点,12:01用am 或者 pm 并没有错误
*/
#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
#define N 2005
using namespace std;
int t,n;
int use_time[N];
int double_time[N];//double_time[i]表示第i个人和第i-1个人一起买票的前
int dp[N];
int main(){
// freopen("in.txt","r",stdin);
scanf("%d",&t);
while(t--){
scanf("%d",&n);
for(int i=;i<=n;i++){
scanf("%d",&use_time[i]);
}
for(int i=;i<=n;i++){
scanf("%d",&double_time[i]);
}
dp[]=;//初始化没有人的时候时间是零
dp[]=use_time[];
for(int i=;i<=n;i++){
//自己买票,还是和前面的人一起买票
dp[i]=min(dp[i-]+use_time[i],dp[i-]+double_time[i]);
}
int h=,m=,s=;
h+=dp[n]/;
m+=dp[n]%/;
s+=dp[n]%;
if(h<) printf("%02d:%02d:%02d am\n",h,m,s);
else printf("%02d:%02d:%02d pm\n",h,m,s);
}
return ;
}
Tickets的更多相关文章
- POJ2828 Buy Tickets[树状数组第k小值 倒序]
Buy Tickets Time Limit: 4000MS Memory Limit: 65536K Total Submissions: 19012 Accepted: 9442 Desc ...
- ACM: FZU 2112 Tickets - 欧拉回路 - 并查集
FZU 2112 Tickets Time Limit:3000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u P ...
- Tickets——H
H. Tickets Jesus, what a great movie! Thousands of people are rushing to the cinema. However, this i ...
- POJ 2828 Buy Tickets(线段树 树状数组/单点更新)
题目链接: 传送门 Buy Tickets Time Limit: 4000MS Memory Limit: 65536K Description Railway tickets were d ...
- 【poj2828】Buy Tickets
Description Railway tickets were difficult to buy around the Lunar New Year in China, so we must get ...
- [poj2828] Buy Tickets (线段树)
线段树 Description Railway tickets were difficult to buy around the Lunar New Year in China, so we must ...
- POJ 2828 Buy Tickets
Description Railway tickets were difficult to buy around the Lunar New Year in China, so we must get ...
- Buy Tickets(线段树)
Buy Tickets Time Limit:4000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u Submit ...
- 【poj2828】Buy Tickets 线段树 插队问题
[poj2828]Buy Tickets Description Railway tickets were difficult to buy around the Lunar New Year in ...
- ural 1217. Unlucky Tickets
1217. Unlucky Tickets Time limit: 1.0 secondMemory limit: 64 MB Strange people live in Moscow! Each ...
随机推荐
- 使用gc、objgraph干掉python内存泄露与循环引用!
Python使用引用计数和垃圾回收来做内存管理,前面也写过一遍文章<Python内存优化>,介绍了在python中,如何profile内存使用情况,并做出相应的优化.本文介绍两个更致命的问 ...
- 基于NIO和BIO的两种服务器对比
基于BIO的服务器,服务端可能要同时保持几百万个HTTP连接,而这些连接并不是每时每刻都在传输数据,所以这种情况不适合使用BIO的服务器:而且需要保证共享资源的同步与安全,这个实现起来相对复杂.这时候 ...
- ssh (免密码登录、开启服务)
ssh 无密码登录要使用公钥与私钥.linux下可以用用ssh-keygen生成公钥/私钥对,下面我以Unbutun为例.有机器A(192.168.1.155),B(192.168.1.181).现想 ...
- hdu 4090--GemAnd Prince(搜索)
题目链接 Problem Description Nowadays princess Claire wants one more guard and posts the ads throughout ...
- Java数据结构和算法总结-冒泡排序、选择排序、插入排序算法分析
前言:排序在算法中的地位自然不必多说,在许多工作中都用到了排序,就像学生成绩统计名次.商城商品销量排名.新闻的搜索热度排名等等.也正因为排序的应用范围如此之广,引起了许多人深入研究它的兴趣,直至今天, ...
- Apache shiro的简单介绍与使用(与spring整合使用)
apache shiro框架简介 Apache Shiro是一个强大而灵活的开源安全框架,它能够干净利落地处理身份认证,授权,企业会话管理和加密.现在,使用Apache Shiro的人越来越多,因为它 ...
- fitnesse - Variables and Symbols
fitnesse - Variables and Symbols 2017-09-30 目录 1 Variables(静态变量) 1.1 定义及使用 1.2 Variable作用域 1.2. ...
- zoj 1889 ones 数学
Ones Time Limit: 2 Seconds Memory Limit: 65536 KB Given any integer 0 <= n <= 10000 not d ...
- 解决Android5.0以下Dialog引起的内存泄漏
最近项目开发中,开发人员和测试人员均反应在android5.0以下手机上LeakCanary频繁监控到内存泄漏,如下图所示,但凡用到Dialog或DialogFragment地方均出现了内存泄漏. 如 ...
- Weave Scope 多主机监控 - 每天5分钟玩转 Docker 容器技术(81)
除了监控容器,Weave Scope 还可以监控 Docker Host. 点击顶部 HOSTS 菜单项,地图将显示当前 host. 与容器类似,点击该 host 图标将显示详细信息. host 当前 ...