1022 Digital Library (30)（30 分）

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1 题目

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

2 解题思路

　　题目大意：一个数字图书馆藏有百万级别的书，书以书名、作者、检索关键字、出版社、出版年月进行存储，每本书对应一个独一无二的7位数字号码作为ID。读者给出一个查询号码，你要输出对应的目标书籍，以升序输出他们的ID信息。
　　这是一个映射问题，一对多的映射，ID可以看做一个主码，其余看做属性，然后通过属性（多个值）反向检索主码（一个值），构成一个std::map的使用场景，因此，可以用map 容器来解决这个问题。

3 AC代码

/*
** @Brief:No.1022 of PAT advanced level.
** @Author:Jason.Lee
** @Date:2018-8-8 
*/
#include 
#include 
#include 
using namespace std;
map<string, set<int> > title, author, key, pub, year;
void query(map<string, set<int> > &m, string &str) {
    if(m.find(str) != m.end()) {
        for(auto it = m[str].begin(); it != m[str].end(); it++)
            printf("%07d\n", *it);
    } else
        cout << "Not Found\n";
}
int main() {
    int n, m, id, num;
    scanf("%d", &n);
    string ttitle, tauthor, tkey, tpub, tyear;
    for(int i = 0; i < n; i++) {
        scanf("%d\n", &id);
        // 书名中可能含有空格，所以要用getline 
        getline(cin, ttitle);
        // 插入到映射中 映射键值id 
        title[ttitle].insert(id);
        getline(cin, tauthor);
        author[tauthor].insert(id);
        // cin不接受换行符，可以作为输入的终点 
        while(cin >> tkey) {
            key[tkey].insert(id);
            char c = getchar();
            if(c == '\n') break;
        }
        getline(cin, tpub);
        pub[tpub].insert(id);
        getline(cin, tyear);
        year[tyear].insert(id);
    }
    scanf("%d", &m);
    for(int i = 0; i < m; i++) {
        scanf("%d: ", &num);
        string temp;
        getline(cin, temp);
        cout << num << ": " << temp << "\n";
        if(num == 1) query(title, temp);
        else if(num == 2) query(author, temp);
        else if(num == 3) query(key, temp);
        else if(num == 4) query(pub,temp);
        else if(num ==5) query(year, temp);
    }
    return 0;
}