Digital Library

1022. Digital Library (30)

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:

31111111The Testing BookYue Chentest code debug sort keywordsZUCS Print20113333333Another Testing BookYue Chentest code sort keywordsZUCS Print220122222222The Testing BookCYLLkeywords debug bookZUCS Print2201161: The Testing Book2: Yue Chen3: keywords4: ZUCS Print5: 20113: blablabla

Sample Output:

1: The Testing Book111111122222222: Yue Chen111111133333333: keywords1111111222222233333334: ZUCS Print11111115: 2011111111122222223: blablablaNot Found

本题要求哦我们能够通过书名或者关键词等匹配书籍，难点在于如何读取并存储给定的信息。

我的做法是：

先用getline()函数读取每一行的字符串

对于key word，我们用stream流再次读取该行字符串，通过空格将每一个keyword存储在数组中

头文件<algorithm>：

可以使用sort()函数对vector进行排序，默认为升序

降序则添加greater参数sort(bookID[i].begin(),bookID[i].end(),greater<string>())，需要包含头文件<functional>

#include <iostream>#include <vector>#include <queue>#include <string>#include <algorithm>
#include <functional>#include <sstream>using namespace std;struct Book {vector<vector<string> > label;Book() {label.resize(6);}};int main() {int n, m;cin >> n;getchar();//skip the enter after 'n' otherwise the getline() while get the '/n'in the first line with 'n';vector<Book> books(n);string ID, title, author, wordlist, publisher, year;string keyword;for (int i = 0;i < n;i++) {getline(cin, ID);getline(cin, title);getline(cin, author);getline(cin, wordlist);getline(cin, publisher);getline(cin, year);books[i].label[0].push_back(ID);books[i].label[1].push_back(title);books[i].label[2].push_back(author);istringstream istr(wordlist);while (!istr.eof()) {istr >> keyword;books[i].label[3].push_back(keyword);}books[i].label[4].push_back(publisher);books[i].label[5].push_back(year);}cin >> m;vector<int> label(m);vector<string> search(m);for (int i = 0;i < m;i++) {cin >> label[i];getchar();getchar();getline(cin, search[i]);}vector<bool> found(m, false);vector<vector<string> > bookID(m);for (int i = 0;i < m;i++) {for (int j = 0;j < n;j++) {for (int k = 0;k < books[j].label[label[i]].size();k++) {if (search[i] == books[j].label[label[i]][k]) {bookID[i].push_back(books[j].label[0][0]);//push the id of the matched bookfound[i] = true;}}}}for (int i = 0;i < m;i++) {cout << label[i] << ": " << search[i] << endl;if (found[i]) {sort(bookID[i].begin(), bookID[i].end());//sort the book ID ascendingfor (int j = 0; j < bookID[i].size(); j++) {cout << bookID[i][j] << endl;}}elsecout << "Not Found" << endl;}system("pause");return 0;}


                                             
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