zoukankan      html  css  js  c++  java
  • pat1022. Digital Library (30)

    1022. Digital Library (30)

    时间限制
    1000 ms
    内存限制
    65536 kB
    代码长度限制
    16000 B
    判题程序
    Standard
    作者
    CHEN, Yue

    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
    

    提交代码

    思路:此题看起来十分麻烦,但是需要仔细分析后思路会更加清晰。 学会利用map 和set的特性,以及二者的组合,比如说map<string,set<int> >一定要记住此处有个空格,其中访问set时需要用迭代器并且只能通过迭代器进行访问,插入时需要用insert。
     1 #include<set>
     2 #include<map>
     3 #include<cstdio>
     4 #include<algorithm>
     5 #include<iostream>
     6 #include<cstring>
     7 #include<queue>
     8 #include<vector>
     9 #include<cmath>
    10 using namespace std;
    11 /*
    12 0: id
    13 1: a book title
    14 2: name of an author
    15 3: a key word
    16 4: name of a publisher
    17 5: a 4-digit number representing the year
    18 */
    19 map<string,set<int> > book[6];
    20 int main(){
    21     //freopen("D:\input.txt","r",stdin);
    22     int n,m;
    23     scanf("%d",&n);
    24     int i,j,id;
    25     string str;
    26     for(i=0;i<n;i++){
    27         scanf("%d",&id);
    28         getchar();//使用cin.getline()特别注意:cin.getline()是读入换行符的 
    29         for(j=1;j<=2;j++){
    30             getline(cin,str);
    31             book[j][str].insert(id);
    32             //cout<<j<<"  "<<str<<endl;
    33             //cout<<book[j][str].size()<<endl;
    34         }
    35         
    36         while(cin>>str){
    37             book[3][str].insert(id);
    38             //cout<<3<<"  "<<str<<endl;
    39             //cout<<book[3][str].size()<<endl;
    40             if(getchar()=='
    '){
    41                 break;
    42             }
    43         }
    44         
    45         
    46         for(j=4;j<=5;j++){
    47             getline(cin,str);
    48             book[j][str].insert(id);
    49             //cout<<j<<"  "<<str<<endl;
    50             //cout<<book[j][str].size()<<endl;
    51         }
    52     }
    53     
    54     scanf("%d",&m);
    55     for(i=0;i<m;i++){
    56         scanf("%d: ",&id);
    57         printf("%d: ",id);
    58         getline(cin,str);
    59         cout<<str<<endl;
    60         if(!book[id].count(str)){
    61             printf("Not Found
    ");
    62             continue;
    63         }
    64         set<int>::iterator it;
    65         for(it=book[id][str].begin();it!=book[id][str].end();it++){
    66             printf("%07d
    ",*it);//cout<<*it<<endl;
    67         }
    68     }
    69     return 0;
    70 }
  • 相关阅读:
    oracle impdp 导入
    oracle权限的分配
    Oracle CASE WHEN 用法介绍
    Oracle自动执行任务(存储过程)
    PL/SQL注册码
    ORACLE基本的sql语句
    ORACLE导出导入问题和表空间问题
    PLSQL笔记
    JSEL 表达式
    asp.net HTTP教程一(HTTP运行期与页面执行模型 )
  • 原文地址:https://www.cnblogs.com/Deribs4/p/4765185.html
Copyright © 2011-2022 走看看