A1022. Digital Library

1022. Digital Library (30)

时间限制

1000 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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: blablabla
Not Found
#include<cstdio>#include<string>#include<iostream>#include<vector>#include<map>#include<utility>using namespace std;map<int, vector<string> > mp;int main(){int n, book_id;scanf("%d", &n);for(int i = 0; i < n; ++i){vector<string> str;   //字符串向量string temp;          //临时字符串scanf("%d", &book_id); getchar();for(int j = 0; j < 5; ++j){getline(cin, temp);str.push_back(temp);}mp.insert(make_pair(book_id, str));}int m, query_id;string info;scanf("%d", &m);getchar();for(int i = 0; i < m; ++i){scanf("%d: ", &query_id);getline(cin, info);cout << query_id << ": " << info << endl;map<int, vector<string> >::iterator it = mp.begin();bool flag = false;if(query_id != 3){            //查找不等于3的情况while(it != mp.end()){if(it->second[query_id - 1] == info){printf("%07d\n", it->first);       //输出书籍idflag = true;}++it;}}else{                //查找等于3的情况while(it != mp.end()){if(it->second[query_id - 1].find(info) != string::npos){printf("%07d\n", it->first);       //输出书籍idflag = true;}++it;}}if(flag == false){printf("Not Found\n");}}return 0;}



这道题也可以用晴神方法，把书按查询类型归为一类，然后建立map<string, set<int> > 将书籍同一类id归为一个集合
例如   map<string, set<int> >  mpTitle, mpAuthor, mpKey, mpPub, mpYear;
不过此题要注意对第三条关键字要注意，录入的时候是整条，但查询的时候是只查询一个关键字，不是像录入那样整条关键字，所以 需要单独拿出来， find()查找
另外值得一提的是，之前用cout输出最后一个测试点是911ms  换成printf最后一个测试点输出时间569ms，可见输出最好用printf，否则很容易超时