A1022. Digital Library
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
1 #include<cstdio> 2 #include<iostream> 3 #include<algorithm> 4 #include<map> 5 #include<string> 6 #include<set> 7 using namespace std; 8 map<string, set<int>> mp1, mp2, mp3, mp4, mp5; 9 void show(map<string, set<int>> &mp, string &key){ 10 set<int> ::iterator it; 11 int find = 0; 12 for(it = mp[key].begin(); it != mp[key].end(); it++){ 13 printf("%07d\n", *it); 14 find = 1; 15 } 16 if(find == 0) 17 printf("Not Found\n"); 18 } 19 int main(){ 20 int N, M, id; 21 string ss, ss2; 22 scanf("%d ", &N); 23 for(int i = 0; i < N; i++){ 24 scanf("%d ", &id); 25 getline(cin, ss); 26 mp1[ss].insert(id); 27 getline(cin, ss); 28 mp2[ss].insert(id); 29 while(cin >> ss){ 30 char c = getchar(); 31 mp3[ss].insert(id); 32 if(c == '\n') 33 break; 34 } 35 getline(cin, ss); 36 mp4[ss].insert(id); 37 getline(cin, ss); 38 mp5[ss].insert(id); 39 } 40 scanf("%d ", &M); 41 for(int i = 0; i < M; i++){ 42 getline(cin, ss2); 43 cout << ss2 + "\n"; 44 ss = ss2.substr(3); 45 if(ss2[0] == '1'){ 46 show(mp1, ss); 47 }else if(ss2[0] == '2'){ 48 show(mp2, ss); 49 }else if(ss2[0] == '3'){ 50 show(mp3, ss); 51 }else if(ss2[0] == '4'){ 52 show(mp4, ss); 53 }else if(ss2[0] == '5'){ 54 show(mp5, ss); 55 } 56 } 57 cin >> N; 58 return 0; 59 }
总结：
1、题意：先输入书的id、名字、作者、关键词、出版社、出版时间。然后根据除书id以外的信息来查询书的id。注意其中书的关键词在输入时为一行字符串，其实是多个关键词以空格分开。由于一个信息可能对多本书（一个出版社有多本书...），且在查询结果时需要有序输出，所以可以使用 map<string , set<int>> 存储。
2、对于下面信息的第4行，需要分别读出每个单词，可以这么做
1111111 The Testing Book Yue Chen test code debug sort keywords ZUCS Print 2011
while(cin >> ss){
char c = getchar();
mp3[ss].insert(id);
if(c == '\n')
break;
}
而当要用cin 读入带空格的一行时，可以 getline(cin ， str)；
3、在函数传参时，如果参数中有map、set、string等，应该传引用，否则可能会超时。
