PAT甲级——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 (≤) 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

#include <iostream>
#include <map>
#include <unordered_map>
#include <set>
#include <string>
using namespace std;

方法一：
struct node
{
    string name, author, keywords, publisher, year;
};
int main()
{
    int N, M;
    string ID;
    cin >> N;
    map<string, node>data;
    for (int i = 0; i < N; ++i)
    {
        node book;
        cin >> ID;
        getchar();//去除回车键
        getline(cin, book.name);
        getline(cin, book.author);
        getline(cin, book.keywords);
        getline(cin, book.publisher);
        cin >> book.year;
        data[ID] = book;
    }
    cin >> M;
    getchar();//去除回车键
    for (int i = 0; i < M; ++i)
    {
        string str;
        bool is = true;
        getline(cin, str);
        cout << str << endl;
        int index = str[0] - '0';
        str.assign(str.begin() + 3, str.end());
        for (auto ptr = data.begin(); ptr != data.end(); ++ptr)
        {
            switch (index)
            {
            case 1:
                if (ptr->second.name == str)
                {
                    is = false;
                    cout << ptr->first << endl;
                }
                break;
            case 2:
                if (ptr->second.author== str)
                {
                    is = false;
                    cout << ptr->first << endl;
                }
                break;
            case 3:
                if (ptr->second.keywords.find(str) !=-1)
                {
                    is = false;
                    cout << ptr->first << endl;
                }
                break;
            case 4:
                if (ptr->second.publisher == str)
                {
                    is = false;
                    cout << ptr->first << endl;
                }
                break;
            case 5:
                if (ptr->second.year == str)
                {
                    is = false;
                    cout << ptr->first << endl;
                }
                break;
            default:
                break;
            }
        }
        if (is)
            cout << "Not Found" << endl;        
    }
    return 0;
}

//方法二
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
{
    if(data.find(str)==data.end())
        printf("Not Found
");
    else
    {
        for (auto ptr = data.find(str)->second.begin(); ptr != data.find(str)->second.end(); ++ptr)
            printf("%07d
", *ptr);
    }
}
int main()
{
    int N, M, ID;
    scanf("%d", &N);
    string til, aut, keys, pub, yea;
    unordered_map<string, set<int>>title, author, keywords, publisher, year;//因为key不唯一
    for (int i = 0; i < N; ++i)
    {
        scanf("%d
", &ID);//不用清除回车键
        getline(cin, til);
        title[til].insert(ID);
        getline(cin, aut);
        author[aut].insert(ID);
        while (cin >> keys)
        {
            keywords[keys].insert(ID);
            char c = getchar();
            if (c == '
')break;
        }
        getline(cin, pub);
        publisher[pub].insert(ID);
        getline(cin, yea);
        year[yea].insert(ID);
    }
    scanf("%d
", &M);
    for (int i = 0; i < M; ++i)
    {
        string str;
        getline(cin, str);
        cout << str << endl;
        int index = str[0] - '0';
        str.assign(str.begin() + 3, str.end());
        if (index == 1) findInfo(title, str);
        else if (index == 2) findInfo(author, str);
        else if (index == 3) findInfo(keywords, str);
        else if (index == 4) findInfo(publisher, str);
        else if (index == 5) findInfo(year, str);
        else printf("Not Found
");
    }
    return 0;
}