• 1022 Digital Library (30 分)(map)


    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
    

    生词

    英文 解释
    query 查询

    题目大意:

    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

    分析:

    1、对除了id之外的其他信息都建立一个map<string, set>,把相应的id插入对应搜索词的map的集合里面,形成一个信息对应一个集合,集合里面是复合条件的书的id

    2、因为对于输入的关键词(可以重复,算是书本对应的tag标签吧~)没有给定关键词的个数,可以使用while(cin >> s)并且判断c = getchar(),c是否等于 ,如果是再退出循环~

    3、建立query,通过传参的形式可以将不同的map名称统一化,先要判断map.find()和m.end()是否相等(或者判断m[str].size()==0?),如果不等再去遍历整个map,输出所有满足条件的id,如果相等就说明不存在这个搜索词对应的id,那么就要输出Not Found~

    4、传参一定要用引用,否则最后一组数据可能会超时~

    原文链接:https://blog.csdn.net/liuchuo/article/details/52263303

    22分题解

    失分分析:

    本来觉得用一个map就行,后来发现如果title、author、key、pub和year如果有相同的值,那么也会出问题。

    举个栗子:

    2
    1111111
    The Testing Book
    Yue Chen
    test code debug sort keywords
    ZUCS Print
    2011
    3333333
    Another Testing Book
    keywords
    test code sort keywords
    ZUCS Print2
    2012
    可以发现第一个的key里的keywords与第二个的author相同,所以查找keywords时两者都会输出,而这样是不对的。

    解决方法

    分别用不同的map记录即可~

    #include <bits/stdc++.h>
    
    using namespace std;
    
    map<string,set<int> > m;
    int main()
    {
    #ifdef ONLINE_JUDGE
    #else
        freopen("1.txt", "r", stdin);
    #endif
        int n,M,id;
        string s;
        scanf("%d",&n);
        for(int i=0;i<n;i++){
            cin>>id;
            //title
            getchar();getline(cin,s);
            //cout<<"title:"<<s<<endl;
            m[s].insert(id);
            //author
            getline(cin,s);
            //cout<<"author:"<<s<<endl;
            m[s].insert(id);
            //key
            while(cin>>s){
                //cout<<"key:"<<s<<endl;
                m[s].insert(id);
                char c=getchar();
                if(c=='
    ') break;
            }
            //pub
            getline(cin,s);
            //cout<<"pub:"<<s<<endl;
            m[s].insert(id);
            //year
            cin>>s;
            //cout<<"year:"<<s<<endl;
            m[s].insert(id);
        }
        scanf("%d",&M);
        for(int i=0;i<M;i++){
            scanf("%d: ",&id);
            getline(cin,s);
            cout<<id<<": "+s<<endl;
            if(m[s].size()==0){
                cout<<s<<endl;
                cout<<"Not Found"<<endl;
                continue;
            }
            set<int>::iterator it;
            for(it=m[s].begin();it!=m[s].end();it++)
                cout<<*it<<endl;
        }
        return 0;
    }
    

    正确题解

    小技巧:
    while(cin>>s){ //cout<<"key:"<<s<<endl; key[s].insert(id); char c=getchar(); if(c==' ') break; }
    多项输入换行结束的判断~

    #include <bits/stdc++.h>
    
    using namespace std;
    
    map<string,set<int> > title, author, key, pub, year;
    void query(map<string,set<int> > &m,string &str){
        //每一个m[str]相当于一个set<int>
        //柳神这里用的是m.find(str) != m.end()判断
        if(m[str].size()!=0){
            //这里就是遍历set的值
            for(auto it=m[str].begin();it!=m[str].end();it++)
                printf("%07d
    ",*it);
        }else printf("Not Found
    ");
    }
    int main()
    {
    #ifdef ONLINE_JUDGE
    #else
        freopen("1.txt", "r", stdin);
    #endif
        int n,M,id;
        string s;
        scanf("%d",&n);
        for(int i=0;i<n;i++){
            cin>>id;
            //title
            getchar();getline(cin,s);
            //cout<<"title:"<<s<<endl;
            title[s].insert(id);
            //author
            getline(cin,s);
            //cout<<"author:"<<s<<endl;
            author[s].insert(id);
            //key
            while(cin>>s){
                //cout<<"key:"<<s<<endl;
                key[s].insert(id);
                char c=getchar();
                if(c=='
    ') break;
            }
            //pub
            getline(cin,s);
            //cout<<"pub:"<<s<<endl;
            pub[s].insert(id);
            //year
            cin>>s;
            //cout<<"year:"<<s<<endl;
            year[s].insert(id);
        }
        scanf("%d",&M);
        for(int i=0;i<M;i++){
            scanf("%d: ",&id);
            getline(cin,s);
            cout<<id<<": "+s<<endl;
            if(id==1) query(title,s);
            else if(id==2) query(author,s);
            else if(id==3) query(key,s);
            else if(id==4) query(pub,s);
            else if(id==5) query(year,s);
        }
        return 0;
    }
    
    

    本文来自博客园,作者:勇往直前的力量,转载请注明原文链接:https://www.cnblogs.com/moonlight1999/p/15522300.html

  • 相关阅读:
    app接口测试
    鼠标右键添加"在此处打开命令窗口"
    解决Maven的Could not resolve archetype org.apache.maven.archetypes:maven-archetype-quickstart
    Linux机器之间复制文件和目录方式&Linux的scp命令详解
    Java动态代理总结
    解决当FORM的ENCTYPE="multipart/form-data" 时request.getParameter()获取不到值的方法
    Elasticsearch使用filter进行匹配关系and,or,not,range查询
    阿里Java开发规范&谷歌Java开发规范&华为Java开发规范&Tab键和空格比较&Eclipse的Tab键设置 总结
    Kafka连接SparkStreaming的两种方式
    IntelliJ IDEA导入多个eclipse项目到同一个workspace下
  • 原文地址:https://www.cnblogs.com/moonlight1999/p/15522300.html
Copyright © 2020-2023  润新知