PAT Advanced 1022 Digital Library (30 分)

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 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 (≤) 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

我们使用一个unordered_map存储即可，unordered_map底层是哈希存储，我们在这题注意2点。

1.Not Found不要忘记

2.unordered_map存储的vector要进行一次排序

#include <iostream>

#include <vector>

#include <sstream>

#include <unordered_map>

#include <algorithm>

using namespace std;

bool cmp(string str1,string str2){

    return str1<str2;

}

int main()

{

    string id,key,tmp1,tmp2;

    unordered_map<string,vector<string>> m;

    //input

    int N;cin>>N;getline(cin,tmp1);

    for(int i=;i<N;i++){

        getline(cin,id);

        getline(cin,key);//book title

        m[key].push_back(id);

        getline(cin,key);//author

        m[key].push_back(id);

        getline(cin,tmp1);//kw

        stringstream ss;

        ss<<tmp1;

        while(ss>>key) m[key].push_back(id);

        getline(cin,key);//pbs

        m[key].push_back(id);

        getline(cin,key);//pbs_y

        m[key].push_back(id);

    }

    int M;cin>>M;getline(cin,tmp1);

    while(M--){//out

        getline(cin,tmp1);

        cout<<tmp1<<endl;

        tmp1=tmp1.substr(,tmp1.length()-);

        sort(m[tmp1].begin(),m[tmp1].end(),cmp);

        if(m[tmp1].size()!=)

            for(int i=;i<m[tmp1].size();i++) cout<<m[tmp1][i]<<endl;

        else cout<<"Not Found"<<endl;

    }

    system("pause");

    return ;

}