PAT 甲级 1022. Digital Library（map，读取标准输入）

原题传送门

• 模拟数字图书馆的查询功能。会给出n本书的信息，以及m个需要查询的命令，数字标号对应相应的命令，数字编号后面的字符串是查询的搜索词，要求输出这行命令以及输出满足条件的书的id，如果一个都没有找到，输出Not Found。

#include <iostream>#include <map>#include <set>using namespace std;map<string, set<int> > title, author, key, pub, year;void query(map<string, set<int> > &m, string &str) { // 传参用引用，防止超时    if(m.find(str) != m.end()) {        for(set<int>::iterator it = m[str].begin(); it != m[str].end(); it++) {            printf("%07d\n", *it);        }    } else {        cout << "Not Found\n";    }}int main() {    int n, id;    cin >> n;    string i_title, i_author, i_key, i_pub, i_year;    // 四种读取标准输入的方式：    for(int i = 0; i < n; ++i) {        cin >> id; // 1.一个int数据        getchar(); // 读取换行符        getline(cin, i_title); // The Testing Book 2.一行数据        title[i_title].insert(id);        getline(cin, i_author); // Yue Chen        author[i_author].insert(id);        while(cin >> i_key) { // test code debug sort keywords 3.可能关键词较多占据多行            key[i_key].insert(id);            char c;            c = getchar();            if(c == '\n') break;        }        getline(cin, i_pub); // ZUCS Print        pub[i_pub].insert(id);        getline(cin, i_year); // 2011        year[i_year].insert(id);    }    int m, num;    cin >> m;    for(int i = 0; i < m; ++i) {        scanf("%d: ", &num); // 4.格式化读取        string temp;        getline(cin, temp);        cout << num << ": " << temp << "\n";        if(num == 1)            query(title, temp);        else if(num == 2)            query(author, temp);        else if(num == 3)            query(key, temp);        else if(num == 4)            query(pub, temp);        else if(num == 5)            query(year, temp);    }    return 0;}

附原题：

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:

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