1022. Digital Library (30)

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:
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

提交代码

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 >。如果是map,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
#include
#include
#include
#include
using namespace std;
map > 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<::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::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::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::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::iterator it = year[temp].begin();it != year[temp].end();it++){
						printf("%07d\n",*it);
					}
				}
				break;
		}
	}
	return 0;
}


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