【PAT】【Advanced Level】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:

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

原题链接：

https://www.patest.cn/contests/pat-a-practise/1022

思路：

按要求存储，对于每个查询请求线性搜索即可

a.find(b)：返回开始位置或者string:npos

CODE:

#include<iostream>#include<cstring>#include<string>#include<vector>#include<cstdio>#include<algorithm>#define N 10010using namespace std;typedef struct Book{  string ID;  string Title;  string Author;  string Key;  string Publisher;  string Year;};Book b[N];bool cmp(Book a,Book b){    return a.ID<b.ID;}int main(){    int n;    cin>>n;    getchar();    for (int i=0;i<n;i++)    {        getline(cin,b[i].ID);        getline(cin,b[i].Title);        getline(cin,b[i].Author);        getline(cin,b[i].Key);        getline(cin,b[i].Publisher);        getline(cin,b[i].Year);    }    sort(b,b+n,cmp);    int m;    cin>>m;    getchar();    for (int i=0;i<m;i++)    {        string t;        getline(cin,t);        cout<<t<<endl;        string ser=t.substr(3,t.length()-3);        int fl=0;        if (t[0]=='1')        {            for (int j=0;j<n;j++)            {                if (b[j].Title==ser)                    cout<<b[j].ID<<endl,fl=1;            }        }        else if(t[0]=='2')        {            for (int j=0;j<n;j++)            {                if (b[j].Author==ser)                    cout<<b[j].ID<<endl,fl=1;            }        }        else if(t[0]=='3')        {            for (int j=0;j<n;j++)            {                if (b[j].Key.find(ser)!=string::npos)                    cout<<b[j].ID<<endl,fl=1;            }        }        else if(t[0]=='4')        {            for (int j=0;j<n;j++)            {                if (b[j].Publisher==ser)                    cout<<b[j].ID<<endl,fl=1;            }        }        else if(t[0]=='5')        {            for (int j=0;j<n;j++)            {                if (b[j].Year==ser)                    cout<<b[j].ID<<endl,fl=1;            }        }        if (fl==0)            cout<<"Not Found"<<endl;    }    return 0;}