1022. Digital Library (30)-PAT甲级真题

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:
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

• 分析
  这道题就是输入每本图书的作者啊名字啊年份啥的信息，还有借书人的id，最后就是让你通过图书各种属性都行，查找到这本书借的人。然后研究了大神的做法，其实就是用map，每一个属性都是一个map，列名是这个属性的值，然后其内容就是一个set集合包含着借阅的人的id。

#include <iostream>#include <cstdio>#include <map>#include <set>#include <string>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 (set<int>::iterator 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_s ("%d" , &n);    string ttitle , tauthor , tkey , tpub , tyear;    for (int i = 0; i < n; i++)    {        scanf_s ("%d" , &id);        getchar ();        getline (cin , ttitle);        title[ttitle].insert (id);        getline (cin , tauthor);        author[tauthor].insert (id);        while (cin >> tkey)        {            key[tkey].insert (id);            char c;            c = getchar ();            if (c == '\n') break;        }        getline (cin , tpub);        pub[tpub].insert (id);        getline (cin , tyear);        year[tyear].insert (id);    }    scanf_s ("%d" , &m);    for (int i = 0; i < m; i++)    {        scanf_s ("%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;}