PAT

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

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给定条件：
1.N本书的属性（编号，名字，作者，关键字，出版社，出版时间）

要求：
1.给定一个属性，输出具有该属性的书的编号

求解：
1.在输出处理的时候，每一个属性对应一个set容器，容器中是具有该属性的书籍编号
2.遍历set容器输出结果

---

#include <iostream>#include <cstdio>#include <map>#include <set>using namespace std;map<string, set<int> > title, author, kw, pub, year;int n, m, id;string ttitle, tauthor, tkw, tpub, tyear, str;void query(map<string, set<int> > &book, string &str) {if(book.find(str) != book.end()) {for(set<int>::iterator it = book[str].begin(); it != book[str].end(); it++) {printf("%07d\n", *it);}} else {printf("Not Found\n");}}int main(){freopen("input.txt", "r", stdin);scanf("%d", &n);for(int i = 0; i < n; i++) {scanf("%d", &id);getchar();getline(cin,ttitle);title[ttitle].insert(id);getline(cin, tauthor);author[tauthor].insert(id);while(cin>>tkw) {kw[tkw].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: ",&n);getline(cin, str);cout<<n<<": "<<str<<endl;if(n == 1) {query(title, str);} else if(n == 2) {query(author, str);} else if(n == 3) {query(kw, str);} else if(n == 4) {query(pub, str);} else if(n == 5) {query(year, str);}}return 0;}