UVa 230 Borrowers(map和set)
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.
"Algorithms" by Sedgewick, R.
"The C Programming Language" by Kernighan, B. and Ritchie, D.
END
BORROW "Algorithms"
BORROW "The C Programming Language"
RETURN "Algorithms"
RETURN "The C Programming Language"
SHELVEEND
Sample Output
Put "The C Programming Language" after "The Canterbury Tales"
Put "Algorithms" after "The C Programming Language"
END
题意
模拟图书馆借书还书(按作者字典序从小到大,再按书名从小到大),BRROW借一本书,RETURN还一本书,SHELVES把书按序一本本放回书架
题解
由于借书还书只给你书名,所以需要用map<string,string>把书名映射到作者,这样才可以用set查找
BRROW借书:就是删除erase一本书的作者和书名
RETURN还书:就是把作者和书名放进set1中
SHELVES放回书架:就是把遍历所有已经还的书it,用lower-bound找到插入后的位子hit,如果当前set为空或者hit==set.begin()就输出first,否则输出hit前面1个(hit--),最后插入*it即可,别忘了最后输出END
代码
#include<bits/stdc++.h>
using namespace std;
struct book
{
string title,author;
book(string t,string a):title(t),author(a){}
bool operator <(const book &rhs)const
{
return author<rhs.author||(author==rhs.author&&title<rhs.title);
}
};
int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
string s;
map<string,string> ma;
set<book> se,se1;
while(getline(cin,s))
{
if(s=="END")break;
int pos=s.find(" by ");
ma[s.substr(,pos)]=s.substr(pos+);
se.insert(book(s.substr(,pos),s.substr(pos+)));
}
while(getline(cin,s))
{
if(s[]=='E')break;
else if(s[]=='S')
{
set<book>::iterator it,bit;
for(it=se1.begin();it!=se1.end();it++)
{
bit=se.lower_bound(*it);
cout<<"Put "<<it->title;
if(se.empty()||bit==se.begin())
cout<<" first"<<endl;
else
cout<<" after "<<(--bit)->title<<endl;
se.insert(*it);
}
se1.clear();
cout<<"END"<<endl;
}
else if(s[]=='B')
se.erase(book(s.substr(),ma[s.substr()]));
else if(s[]=='R')
se1.insert(book(s.substr(),ma[s.substr()]));
}
return ;
}
UVa 230 Borrowers(map和set)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
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: ...
- Borrowers UVA - 230
I mean your borrowers of books - those mutilators of collections, spoilers of the symmetry of shel ...
- 【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 156 Ananagrams ---map
题目链接 题意:输入一些单词,找出所有满足如下条件的单词:该单词不能通过字母重排,得到输入文本中的另外一个单词.在判断是否满足条件时,字母不分大小写,但在输出时应保留输入中的大小写,按字典序进行排列( ...
- uva 156 (map)
暑假培训习题 1.用vector<string>储存string类型的输入单词: 2.将vector中的元素逐一标准化后映射进map中,并给map值加一: 3.新建一个空的vector 4 ...
- UVA 12035 War Map
让我先把微笑送给出题人 这个题最基础的一个想法:先找出一个度数和为总度数和的1/2的点集,然后判断这个点集和这个点集的补集能否形成二分图.但是就算我们把判断的复杂度看成O(1),这个算法的复杂度也是 ...
- UVa 1592 Database (map)
题意:给出n行m列的数据库(数据范围: n 1~10000, m 1~10), 问你能不能找出两行r1, r2,使得这两行中的c1, c2列是一样的, 即(r1,c1)==(r2,c1) && ...
随机推荐
- pandas的merge函数
pandas.merge(left,right,how='inner',on=None,left_on=None,right_on=None,left_index=False,right_index= ...
- SQL查询日期:
SQL查询日期: 今天的所有数据:select * from 表名 where DateDiff(dd,datetime类型字段,getdate())=0 昨天的所有数据:select * from ...
- unity开发android游戏
环境搭建: Unity+JDK+Android Studio+Android SDK(+NDK) 教程:unity开发android游戏(一)搭建Unity安卓开发环境 注意“Build System ...
- python中os常用方法
python中OS常用方法 Python的标准库中的os模块包含普遍的操作系统功能.如果你希望你的程序能够与平台无关的话,这个模块是尤为重要的.即它允许一个程序在编写后不需要任何改动,也不会发生任何问 ...
- Spark SQL 基本原理
Spark SQL 模块划分 Spark SQL架构--catalyst设计图 Spark SQL 运行架构 Hive的兼容性
- 信息学奥赛(NOIP)初赛学习方法推荐
首先声明:本帖针对初学者,本帖只是列出一个大概的框架,不属于自学方法,有条件有能力,请找一位好老师来教,多跟前辈交流经验.(否则多会出现事倍功半的悲剧!) 一.初赛内容 初赛偏重于基础知识. 1. 一 ...
- blktrace未公开选项网络保存截取数据
本文链接地址: blktrace未公开选项网络保存截取数据 我们透过blktrace来观察io行为的时候,第一件事情需要选择目标设备,以便分析该设备的io行为.具体使用可以参考我之前写的几篇:这里 这 ...
- mysql更新(七) MySQl创建用户和授权
14-补充内容:MySQl创建用户和授权 权限管理 我们知道我们的最高权限管理者是root用户,它拥有着最高的权限操作.包括select.update.delete.update.grant等操作 ...
- jQuery实现todo及轮播图
内容: 1.todo程序 2.轮播图 1.todo程序 需求: 实现一个todo程序,可以添加数据,可以删除数据,可以修改数据,可以查看所有数据 另外实现自己的一系列弹窗:用于提示用户的提示框.用于警 ...
- scrapy与redis分布式组件
Scrapy 和 scrapy-redis的区别 Scrapy 是一个通用的爬虫框架,但是不支持分布式,Scrapy-redis是为了更方便地实现Scrapy分布式爬取,而提供了一些以redis为基础 ...