• 热门专题

1022.DigitalLibrary(30)PAT(AdvancedLevel)Practise

作者:  发布日期:2015-12-02 20:44:02
  • 题目信息

    1022. Digital Library (30)

    时间限制1000 ms
    内存限制65536 kB
    代码长度限制16000 B

    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

    解题思路

    用map和set组合存储数据,规避排序等问题

    AC代码

    #include <cstdio>
    #include <map>
    #include <set>
    #include <string>
    #include <cstring>
    using namespace std;
    
    map<string, set<string> > bookName;
    map<string, set<string> > authorName;
    map<string, set<string> > keywords;
    map<string, set<string> > publisher;
    map<string, set<string> > year;
    
    void out(map<string, set<string> > &mp, char *s){
        if (mp[s].size() > 0){
            for (set<string>::iterator it = mp[s].begin(); it != mp[s].end(); ++it){
                printf("%s\n", it->c_str());
            }
        }else{
            printf("Not Found\n");
        }
    }
    int main()
    {
        char sid[100], s[100], ts[100];
        int n, a;
        scanf("%d", &n);
        gets(s);
        while (n--){
            gets(sid);
            gets(s);
            bookName[s].insert(sid);
            gets(s);
            authorName[s].insert(sid);
            gets(s);
            char *p = strtok(s, " ");
            while (p){
                keywords[p].insert(sid);
                p = strtok(NULL, " ");
            }
            gets(s);
            publisher[s].insert(sid);
            gets(s);
            year[s].insert(sid);
        }
        scanf("%d", &n);
        gets(s);
        while (n--){
            scanf("%d: ", &a);
            gets(s);
            printf("%d: %s\n", a, s);
            if (a == 1) out(bookName, s);
            else if (a == 2) out(authorName, s);
            else if (a == 3) out(keywords, s);
            else if (a == 4) out(publisher, s);
            else if (a == 5) out(year, s);
        }
        return 0;
    }

延伸阅读:

About IT165 - 广告服务 - 隐私声明 - 版权申明 - 免责条款 - 网站地图 - 网友投稿 - 联系方式
本站内容来自于互联网,仅供用于网络技术学习,学习中请遵循相关法律法规