1022 Digital Library (30)（30 分）
A Digital Library contains millions of books, stored according to their titles, authors, key words of their abstracts, publishers, and published years. Each book is assigned an unique 7-digit number as its ID. Given any query from a reader, you are supposed to output the resulting books, sorted in increasing order of their ID's.
Input Specification:
Each input file contains one test case. For each case, the first line contains a positive integer N (<=10000) which is the total number of books. Then N blocks follow, each contains the information of a book in 6 lines:
- Line #1: the 7-digit ID number;
- Line #2: the book title -- a string of no more than 80 characters;
- Line #3: the author -- a string of no more than 80 characters;
- Line #4: the key words -- each word is a string of no more than 10 characters without any white space, and the keywords are separated by exactly one space;
- Line #5: the publisher -- a string of no more than 80 characters;
- Line #6: the published year -- a 4-digit number which is in the range [1000, 3000].
It is assumed that each book belongs to one author only, and contains no more than 5 key words; there are no more than 1000 distinct key words in total; and there are no more than 1000 distinct publishers.
After the book information, there is a line containing a positive integer M (<=1000) which is the number of user's search queries. Then M lines follow, each in one of the formats shown below:
- 1: a book title
- 2: name of an author
- 3: a key word
- 4: name of a publisher
- 5: a 4-digit number representing the year
Output Specification:
For each query, first print the original query in a line, then output the resulting book ID's in increasing order, each occupying a line. If no book is found, print "Not Found" instead.
Sample Input:
3
1111111
The Testing Book
Yue Chen
test code debug sort keywords
ZUCS Print
2011
3333333
Another Testing Book
Yue Chen
test code sort keywords
ZUCS Print2
2012
2222222
The Testing Book
CYLL
keywords debug book
ZUCS Print2
2011
6
1: The Testing Book
2: Yue Chen
3: keywords
4: ZUCS Print
5: 2011
3: blablabla
Sample Output:
1: The Testing Book
1111111
2222222
2: Yue Chen
1111111
3333333
3: keywords
1111111
2222222
3333333
4: ZUCS Print
1111111
5: 2011
1111111
2222222
3: blablabla
Not Found题目大意：输入一个整数n,接下来共有n本书输入，书包括几个属性；接下来再进行查询，输出对应的id.
代码来自:https://www.liuchuo.net/archives/2295
#include <iostream>
#include <map>
#include <set>
using namespace std;
map<string, set<int> > title, author, key, pub, year;
void query(map<string, set<int> > &m, string &str) {//需要使用引用传参，不然就超时了。
if(m.find(str) != m.end()) {//find寻找是否有这个关键字。
for(auto it = m[str].begin(); it != m[str].end(); it++)
printf("%07d\n", *it);//这里需要是07，因为
} else
cout << "Not Found\n";
}
int main() {
int n, m, id, num;
scanf("%d", &n);
string ttitle, tauthor, tkey, tpub, tyear;
for(int i = ; i < n; i++) {
scanf("%d\n", &id);
getline(cin, ttitle);
title[ttitle].insert(id);//title是互异的，插入对应set中的id。
getline(cin, tauthor);//getline可以直接读入到string里啊。
author[tauthor].insert(id);//这个author下有多少个
while(cin >> tkey) {//解决有空格的问题。
key[tkey].insert(id);//这个关键词下对应的id。
char c = getchar();//正常会获取到空格。
if(c == '\n') break;
}
getline(cin, tpub);
pub[tpub].insert(id);
getline(cin, tyear);
year[tyear].insert(id);
}
scanf("%d", &m);
for(int i = ; i < m; i++) {
scanf("%d: ", &num);//一行内还可以分开读入。
string temp;
getline(cin, temp);
cout << num << ": " << temp << "\n";
if(num == ) query(title, temp);
else if(num == ) query(author, temp);
else if(num == ) query(key, temp);
else if(num == ) query(pub,temp);
else if(num ==) query(year, temp);
}
return ;
}
//简直不要太厉害啊，我的思路是，建立5个map，不过每个map的关键字都是id,每个id都对应一些结构，然后查询的时候，遍历每个id，判断其下是否存在对应的查询条件，这样真的很慢啊；还有对与使用getline时遇到了问题。这个代码和我的思路正好相反，十分简洁。
1.可以使用getline(cin,str);实现对string的输入；
2.使用getchar()可以获取多余的空格或者换行符；
3.传参时使用引用，可以减少调用时间；
4.scanf读入时的特点，map.find()中查找关键字是否存在；
5.对于输入的内容又输出，像代码中的for循环，直接操作之后输出，不用中间存储。