PAT甲级——A1022 Digital Library

Posted 2022-05-16 zzw1024

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 (≤) which is the total number of books. Then Nblocks 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 (≤) 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

第一种方法
以时间换空间，就是个ID保留一组信息，查询时遍历查询
但可能导致时间复杂度过大

第二种方法[推荐使用方法二]
以空间换时间，每种信息对应一个ID，查找时，时间复杂度为O(1)
但可能导致空间复杂度太大

注意一些字符输入的细节

 

  1 #include <iostream>
  2 #include <map>
  3 #include <unordered_map>
  4 #include <set>
  5 #include <string>
  6 using namespace std;
  7 
  8 方法一：
  9 struct node
 10 
 11     string name, author, keywords, publisher, year;
 12 ;
 13 int main()
 14 
 15     int N, M;
 16     string ID;
 17     cin >> N;
 18     map<string, node>data;
 19     for (int i = 0; i < N; ++i)
 20     
 21         node book;
 22         cin >> ID;
 23         getchar();//去除回车键
 24         getline(cin, book.name);
 25         getline(cin, book.author);
 26         getline(cin, book.keywords);
 27         getline(cin, book.publisher);
 28         cin >> book.year;
 29         data[ID] = book;
 30     
 31     cin >> M;
 32     getchar();//去除回车键
 33     for (int i = 0; i < M; ++i)
 34     
 35         string str;
 36         bool is = true;
 37         getline(cin, str);
 38         cout << str << endl;
 39         int index = str[0] - ‘0‘;
 40         str.assign(str.begin() + 3, str.end());
 41         for (auto ptr = data.begin(); ptr != data.end(); ++ptr)
 42         
 43             switch (index)
 44             
 45             case 1:
 46                 if (ptr->second.name == str)
 47                 
 48                     is = false;
 49                     cout << ptr->first << endl;
 50                 
 51                 break;
 52             case 2:
 53                 if (ptr->second.author== str)
 54                 
 55                     is = false;
 56                     cout << ptr->first << endl;
 57                 
 58                 break;
 59             case 3:
 60                 if (ptr->second.keywords.find(str) !=-1)
 61                 
 62                     is = false;
 63                     cout << ptr->first << endl;
 64                 
 65                 break;
 66             case 4:
 67                 if (ptr->second.publisher == str)
 68                 
 69                     is = false;
 70                     cout << ptr->first << endl;
 71                 
 72                 break;
 73             case 5:
 74                 if (ptr->second.year == str)
 75                 
 76                     is = false;
 77                     cout << ptr->first << endl;
 78                 
 79                 break;
 80             default:
 81                 break;
 82             
 83         
 84         if (is)
 85             cout << "Not Found" << endl;        
 86     
 87     return 0;
 88 
 89 
 90 //方法二
 91 void findInfo(unordered_map<string, set<int>>&data,string &str)//传参一定要用引用，否则最后一组数据可能会超时
 92 
 93     if(data.find(str)==data.end())
 94         printf("Not Found\n");
 95     else
 96     
 97         for (auto ptr = data.find(str)->second.begin(); ptr != data.find(str)->second.end(); ++ptr)
 98             printf("%07d\n", *ptr);
 99     
100 
101 int main()
102 
103     int N, M, ID;
104     scanf("%d", &N);
105     string til, aut, keys, pub, yea;
106     unordered_map<string, set<int>>title, author, keywords, publisher, year;//因为key不唯一
107     for (int i = 0; i < N; ++i)
108     
109         scanf("%d\n", &ID);//不用清除回车键
110         getline(cin, til);
111         title[til].insert(ID);
112         getline(cin, aut);
113         author[aut].insert(ID);
114         while (cin >> keys)
115         
116             keywords[keys].insert(ID);
117             char c = getchar();
118             if (c == ‘\n‘)break;
119         
120         getline(cin, pub);
121         publisher[pub].insert(ID);
122         getline(cin, yea);
123         year[yea].insert(ID);
124     
125     scanf("%d\n", &M);
126     for (int i = 0; i < M; ++i)
127     
128         string str;
129         getline(cin, str);
130         cout << str << endl;
131         int index = str[0] - ‘0‘;
132         str.assign(str.begin() + 3, str.end());
133         if (index == 1) findInfo(title, str);
134         else if (index == 2) findInfo(author, str);
135         else if (index == 3) findInfo(keywords, str);
136         else if (index == 4) findInfo(publisher, str);
137         else if (index == 5) findInfo(year, str);
138         else printf("Not Found\n");
139     
140     return 0;
141