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: blablablaNot Found

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题目意思：

给出N本书的编号、书名、作者、关键字、出版社、出版年份，随后M各查询，每个查询给出书名、作者、关键字、出版社、出版年份中的一个，根据查询条件输出查询到的所有书的编号。

解题思路：

根据题目意思可以看出，一个字符串可能对应多个编号,换句话说，一个字符串可能对应一个编号数组。那么我们就可以使用map来处理。map<string,set<int> >。如果是map<string,int>，表示一个字符串映射到一个整数上，如map["helllo"] = "5";该题我们直接可以将一个字符串映射到一个数组中，因为set会自动帮助我们排序，所以选用set数组。

下面是具体代码：

#include<iostream>#include<cstring>#include<string>#include<map>#include<set>using namespace std;map<string,set<int> > title,author,key,pub,year;int main(){int n,m,id,type;//freopen("input.txt","r",stdin);string title1,author1,key1,pub1,year1;scanf("%d",&n);while(n--){scanf("%d",&id);char ch = getchar();getline(cin,title1);title[title1].insert(id);getline(cin,author1);author[author1].insert(id);while(cin>>key1){key[key1].insert(id);ch = getchar();if(ch == '\n'){break;}}getline(cin,pub1);pub[pub1].insert(id);getline(cin,year1);year[year1].insert(id);}string temp;scanf("%d",&n);while(n--){scanf("%d: ",&type);getline(cin,temp);cout<<type<<": "<<temp<<endl;switch(type){case 1:if(title.find(temp)==title.end()){printf("Not Found\n");}else{for(set<int>::iterator it = title[temp].begin();it != title[temp].end();it++){printf("%07d\n",*it);}}break;   case 2:   if(author.find(temp)==author.end()){printf("Not Found\n");}else{for(set<int>::iterator it = author[temp].begin();it != author[temp].end();it++){printf("%07d\n",*it);}}break;   case 3:   if(key.find(temp)==key.end()){printf("Not Found\n");}else{for(set<int>::iterator it = key[temp].begin();it != key[temp].end();it++){printf("%07d\n",*it);}}break;   case 4:   if(pub.find(temp)==pub.end()){printf("Not Found\n");}else{for(set<int>::iterator it = pub[temp].begin();it != pub[temp].end();it++){printf("%07d\n",*it);}}break;   case 5:   if(year.find(temp)==year.end()){printf("Not Found\n");}else{for(set<int>::iterator it = year[temp].begin();it != year[temp].end();it++){printf("%07d\n",*it);}}break;}}return 0;}