PAT A1022 Digital Library

1022 Digital Library

分数 30

作者 CHEN, Yue

单位浙江大学

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 (≤104) 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

* 由于题目的输出由查询编号决定，那么就可以将每个查询编号对应的信息用一个map容器
* 存储，map的第一个值的类型为string，存储每个序号对应的信息（书名，作者，...）的
* 字符串，但是输入的关键词要将其拆分成单词存储，因为查询的关键词是以单词进行查询的；
* map的第二个值得类型为set<string>,存储每个信息所对应的书的编号，由于set是自动升序
* 排序的，这也正符合题目的要求；
*
* 最后要注意查询时的输入，我开始用的是两个string来进行输入，即：
* string op, t;
* cin >> op;
* getline(cin, t);
* 但要注意此时的t是连中间的那个空格也一起读入的，中间用getchar()来吸收掉那个空格应该是
* 可以的；
*
* 其实完全可以直接这样写，更加便捷
* getline(cin, op);
    t = op.substr(3);

    map就是一个二维数组，当然可以这么写了：
    for(auto &a : mp[d][t]) mp是定义的一个数组，相当于此时的mp是三维数组
        cout << a << endl;

/**
 * 由于题目的输出由查询编号决定，那么就可以将每个查询编号对应的信息用一个map容器
 * 存储，map的第一个值的类型为string，存储每个序号对应的信息（书名，作者，...）的
 * 字符串，但是输入的关键词要将其拆分成单词存储，因为查询的关键词是以单词进行查询的；
 * map的第二个值得类型为set<string>,存储每个信息所对应的书的编号，由于set是自动升序
 * 排序的，这也正符合题目的要求；
 * 
 * 最后要注意查询时的输入，我开始用的是两个string来进行输入，即：
 *  string op, t;
 *  cin >> op;
 *  getline(cin, t);
 *  但要注意此时的t是连中间的那个空格也一起读入的，中间用getchar()来吸收掉那个空格应该是
 *  可以的；
 * 
 *  其实完全可以直接这样写，更加便捷
 *  getline(cin, op);
    t = op.substr(3);
    
    map就是一个二维数组，当然可以这么写了：
    for(auto &a : mp[d][t]) mp是定义的一个数组，相当于此时的mp是三维数组
        cout << a << endl;
*/

#include <iostream>
#include <map>
#include <set>
#include <string>

using namespace std;

map<string, set<string> > mp[6];

void Read()
{
    int n;
    cin >> n;
    getchar();  //吸收掉换行符必不可少
    string s[7];
    
    for(int i=0; i<n; ++i)
    {
        for(int j=0; j<6; ++j)
        {
            getline(cin, s[j]);
            if(j > 0 && j != 3) //第四个字符串需要特别处理一下，将其分解成单词
                mp[j][s[j]].insert(s[0]);
        }
        
        //第四个字符串需要特别处理一下，将其分解成单词
        int idx = 0;
        while(idx < s[3].size())
        {
            int len = 0, sta = idx;
            while(idx < s[3].size() && s[3][idx] != ' ')
                ++len , ++idx;
            string t = s[3].substr(sta, len);
            mp[3][t].insert(s[0]);
            ++idx;
        }
    }
    
    int m;
    cin >> m;
    getchar(); //吸收掉换行符
    
    string op, t;
    while (m -- )
    {
        getline(cin, op);
        t = op.substr(3);
        cout << op << endl;
        int d = op[0] - '0';
        /**
        scanf("%s ",op); //要这么写就需要将op改为字符数组
        getline(cin, t);
        int d = op[0] - '0';
        cout << op << ' ' << t << endl;
        */
        
        if(mp[d].find(t) == mp[d].end())
            puts("Not Found");
        else
        {
            for(auto &a : mp[d][t])
                cout << a << endl;
        }
            
    }
}

int main()
{
    Read();
    
    return 0;
}