Borrowers UVA - 230
I mean your borrowers of books — those mutilators of collections, spoilers of the symmetry of shelves, and creators of odd volumes.
– (Charles Lamb, Essays of Elia (1823) ‘The Two Races of Men’)
Like Mr. Lamb, librarians have their problems with borrowers too. People don’t put books back where they should. Instead, returned books are kept at the main desk until a librarian is free to replace them in the right places on the shelves. Even for librarians, putting the right book in the right place can be very time-consuming. But since many libraries are now computerized, you can write a program to help.
When a borrower takes out or returns a book, the computer keeps a record of the title. Periodically, the librarians will ask your program for a list of books that have been returned so the books can be returned to their correct places on the shelves. Before they are returned to the shelves, the returned books are sorted by author and then title using the ASCII collating sequence. Your program should output the list of returned books in the same order as they should appear on the shelves. For each book, your program should tell the librarian which book (including those previously shelved) is already on the shelf before which the returned book should go.
Input
First, the stock of the library will be listed, one book per line, in no particular order. Initially, they are all on the shelves. No two books have the same title. The format of each line will be:
"title" by author
The end of the stock listing will be marked by a line containing only the word:
END
Following the stock list will be a series of records of books borrowed and returned, and requests from librarians for assistance in restocking the shelves. Each record will appear on a single line, in one of the following formats:
BORROW title
RETURN title
SHELVE
The list will be terminated by a line containing only the word:
END
Output
Each time the SHELVE command appears, your program should output a series of instructions for the librarian, one per line, in the format:
Put title1 after title2 or, for the special case of the book being the first in the collection:
Put title first
After the set of instructions for each SHELVE, output a line containing only the word:
END
Assumptions & Limitations:
1. A title is at most 80 characters long.
2. An author is at most 80 characters long.
3. A title will not contain the double quote (") character.
Sample Input
"The Canterbury Tales" by Chaucer, G.
"The Canterbury Taless" by Chaucer, B.
"Algorithms" by Sedgewick, R.
"The C Programming Language" by Kernighan, B. and Ritchiee, D.
"The C Programming Languag" by Kernighan, B. and Ritchiee, D.
"The D Programming Language" by Kernighan, B. and Ritchiee, D.
"A House for Mr. Biswas" by Naipaul, V.S.
"A Congo Diary" by Naipaul, V.S.
END
BORROW "Algorithms"
BORROW "The C Programming Language"
BORROW "The C Programming Languag"
BORROW "The Canterbury Taless"
SHELVE
RETURN "Algorithms"
RETURN "The Canterbury Taless"
SHELVE
RETURN "The C Programming Languag"
SHELVE
BORROW "The C Programming Languag"
BORROW "The Canterbury Taless"
BORROW "A House for Mr. Biswas"
RETURN "The Canterbury Taless"
SHELVE
RETURN "The C Programming Language"
RETURN "A House for Mr. Biswas"
SHELVE
END
Sample Output
Put "The C Programming Language" after "The Canterbury Tales"
Put "Algorithms" after "The C Programming Language"
END
HINT
使用结构体数组进行操作。定义一个结构体,成员为头作者和书名两个字符串。排序使用sort(),但要自己定义排序方法。另外,借出书籍的时候,要想定位并删除此元素要自己定义一个指针,使用find()函数无法对结构体进行比较,在结构体里面重载运算符应该是可以的,但小白表示不会。。。所以定义一个auto指针来查找。因为输出书名需要双引号所以存入就不去除双引号了。
整体的框架很好理解,录入书籍整行读取,然后分割书名和作者。然后读入指令,对指令区分操作,小细节还是比较多的,具体看代码。
#include<bits/stdc++.h>
using namespace std;
struct books {
string title; //书名
string name; //作者
};
bool compare(books a, books b) { //自定义的排序方法,用于sort
if (a.name == b.name)return a.title < b.title;
return a.name < b.name;
}
int main(){
vector<books>list; //目前书架上的书籍
vector<books>ret; //归还的书籍
map<string, string>alllist; //存储书籍副本
string s,title;books temp;
while (getline(cin, s) && s != "END") { //录入书籍信息
int i=s.find('\"',1); //对书籍进行分割
temp.title = s.substr(0, i+1);
temp.name = s.substr(i + 1, s.size() - 1);
list.push_back(temp); //存入数组并且保存map副本
alllist[temp.title] = temp.name;
}
while (cin>>s&&s!="END"){
if (s == "SHELVE") { //对未上架的图书上架处理
sort(ret.begin(), ret.end(), compare); //排序
for (int i = 0;i < ret.size();i++) { //挨个书籍处理
temp = ret[i];
list.push_back(temp); //上架
sort(list.begin(), list.end(), compare);//排序
for (int i = 0;i < list.size();i++) { //查找并输出
if (list[i].name == temp.name && list[i].title == temp.title) {
if(i==0)cout << "Put " << temp.title << " first" << endl;
else cout << "Put " << temp.title << " after " << list[i-1].title << endl;
break;
}
}
}
ret.clear(); //每次输出完毕必须清空记录,一面影响下一次的使用
cout << "END" << endl;
}
else {
getchar(); //吃掉空格
getline(cin, title);//读入标题
temp.title = title; //保存到临时结构体
temp.name = alllist[title];
if (s == "BORROW") {
auto pointer = list.begin();//本来使用find()函数不会重载所以自己定义指针查找
while (!((*pointer).title == temp.title))pointer++;
list.erase(pointer); //删除
}
else if (s == "RETURN")ret.push_back(temp);//记录还书信息
}
}
}
Borrowers UVA - 230的更多相关文章
- uva 230 Borrowers(摘)<vector>"结构体“ 膜拜!
I mean your borrowers of books--those mutilators of collections, spoilers of the symmetry of shelves ...
- UVa 230 Borrowers(map和set)
I mean your borrowers of books - those mutilators of collections, spoilers of the symmetry of shelve ...
- Uva - 230 - Borrowers
AC代码: #include <iostream> #include <cstdio> #include <cstdlib> #include <cctype ...
- UVA 230 Borrowers (STL 行读入的处理 重载小于号)
题意: 输入若干书籍和作者名字,然后先按作者名字升序排列,再按标题升序排列,然后会有3种指令,BORROW,RETURN, SHELVE. BORROW 和 RETURN 都会带有一个书名在后面,如: ...
- 【习题 5-8 UVA - 230】Borrowers
[链接] 我是链接,点我呀:) [题意] 在这里输入题意 [题解] 用map+set写个模拟就好. 3个区域 书架.桌子.别人的手上. 其中前两个区域的书都能借出去. [代码] #include &l ...
- UVa 1640 (计数) The Counting Problem
题意: 统计[a, b]或[b, a]中0~9这些数字各出现多少次. 分析: 这道题可以和UVa 11361比较来看. 同样是利用这样一个“模板”,进行区间的分块,加速运算. 因为这里没有前导0,所以 ...
- UVA 10194 (13.08.05)
:W Problem A: Football (aka Soccer) The Problem Football the most popular sport in the world (ameri ...
- UVa 1640 - The Counting Problem(数论)
链接: https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem& ...
- Fast Matrix Operations(UVA)11992
UVA 11992 - Fast Matrix Operations 给定一个r*c(r<=20,r*c<=1e6)的矩阵,其元素都是0,现在对其子矩阵进行操作. 1 x1 y1 x2 y ...
随机推荐
- 物联网网关开发:基于MQTT消息总线的设计过程(上)
道哥的第 021 篇原创 目录 一.前言 二.网关的作用 2.1 指令转发 2.2 外网通信 2.3 协议转换 2.4 设备管理 2.5 边沿计算(自动化控制) 三.网关内部进程之间的通信 3.1 网 ...
- 将项目加载到tomcat中的时候报错:Tomcat version 6.0 only supports J2EE 1.2, 1.3, 1.4, and Java EE 5 Web modules
转自:http://jingwang0523.blog.163.com/blog/static/9090710320113294551497/ 最近在用eclipse做项目,新建项目时什么都贪新,用最 ...
- SQL学习笔记——创建数据库显示:文件激活错误,物理文件名不存在>>解决方案
今天在创建数据库时,跟着老师一步一步的操作创建成功,但出于在厌恶冗长的数据库存储路径,于是,擅自更改了数据filename,让他保存在电脑桌面新建的文件夹,可是一执行就报错了. 老师源码: 1 cre ...
- MVC base64加密的文件,前端下载
后端代码: public FileResult OutPutFile(string base64file,string filename) { buffer = Convert.FromBase64 ...
- CVE-2019-12409-Apache Solr JMX服务远程代码执行
漏洞分析 https://www.freebuf.com/vuls/218730.html 漏洞介绍 该漏洞源于默认配置文件solr.in.sh中的ENABLE_REMOTE_JMX_OPTS配置选项 ...
- 鸿蒙OS前端开发入门指南:网络图片_Image渲染网络图片
目录: 1.开启明文传输 2.权限申请 3.引入http插件 4.案例展示 5.<鸿蒙OS前端开发入门指南>文章合集 开启明文传输 在config.json配置文件添加如下配置(如果不开启 ...
- 顺序表及基本操作(C语言)
#include <stdio.h> #include <stdlib.h> //基本操作函数用到的状态码 #define TRUE 1; #define FALSE 0; # ...
- Django之模版层
一.模版简介 你可能已经注意到我们在例子视图中返回文本的方式有点特别,也就是说,HTML被直接硬编码在python代码之中. def current_datetime(request): now = ...
- 程序一直处于Accept状态,无法调度运行
问题描述:在现场或测试环境偶尔会出现用户提交的程序一直处于Accept状态无法调度运行的现象 问题分析:出现这种问题的原因一般有以下两种: 1.用户程序提交的队列当前是否已达到最大可运行程序数,当达到 ...
- 如何使用Typora写博客
如何写博客及Typora的使用 Typora Typora是写好博客的一个重要的软件,下面我们来介绍如何安装以及使用它 安装 官网下载Typora 较慢,首先附上Typora安装包: 链接:https ...