problem

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

tip

  • 考察map的使用。

answer

#include<iostream>
#include<string>
#include<map>
#include<set>
#include<vector>
using namespace std; int N, M;
map<string, set<int> > titleM, authorM, keyM, pubM, yearM; vector<string> Explode(string s){
string buff = "";
vector<string> v;
for(int i = 0; i < s.size(); i++){
if(s[i] != ' ') buff += s[i];
if(s[i] == ' ' && buff != ""){
v.push_back(buff);
buff = "";
}
}
if(buff != "") v.push_back(buff);
return v;
} void Query(string &q, map<string, set<int> > &m){
set<int>::iterator itS;
if(m.find(q) != m.end())
for(itS = m[q].begin(); itS != m[q].end(); itS++){
printf("%07d\n", *itS);
}
else cout<<"Not Found"<<endl;
} int main(){
// freopen("test.txt", "r", stdin);
scanf("%d", &N);
for(int i = 0; i < N; i ++){
int id;
string title, author, key, pub, year;
scanf("%d\n", &id);
getline(cin, title);
getline(cin, author);
getline(cin, key);
getline(cin, pub);
getline(cin, year); titleM[title].insert(id);
authorM[author].insert(id);
pubM[pub].insert(id);
yearM[year].insert(id);
vector<string> keys = Explode(key);
for(int j = 0; j < keys.size(); j++){
keyM[keys[j]].insert(id);
}
}
scanf("%d", &M);
for(int i = 0; i < M; i++){
int num;
scanf("%d: ", &num);
string temp;
getline(cin, temp);
cout<<num<<": "<<temp<<endl;
switch(num){
case 1:{
Query(temp, titleM);
break;
}
case 2:{
Query(temp, authorM);
break;
}
case 3:{
Query(temp, keyM);
break;
}
case 4:{
Query(temp, pubM);
break;
}
case 5:{
Query(temp, yearM);
break;
}
}
}
return 0;
}

exprience

  • 熟练使用getline(), getchar(), scanf() 与printf()

  • 需要特别注意题目中出现的带有位数描述的数字,再输出时需要按位数输出。

1022 Digital Library (30)(30 point(s))的更多相关文章

  1. PAT 甲级 1022 Digital Library (30 分)(字符串读入getline,istringstream,测试点2时间坑点)

    1022 Digital Library (30 分)   A Digital Library contains millions of books, stored according to thei ...

  2. pat 甲级 1022. 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 Di ...

  3. 1022 Digital Library (30 分)

    1022 Digital Library (30 分)   A Digital Library contains millions of books, stored according to thei ...

  4. PAT 1022 Digital Library[map使用]

    1022 Digital Library (30)(30 分) A Digital Library contains millions of books, stored according to th ...

  5. PTA 1004 Counting Leaves (30)(30 分)(dfs或者bfs)

    1004 Counting Leaves (30)(30 分) A family hierarchy is usually presented by a pedigree tree. Your job ...

  6. A1095 Cars on Campus (30)(30 分)

    A1095 Cars on Campus (30)(30 分) Zhejiang University has 6 campuses and a lot of gates. From each gat ...

  7. 1022 Digital Library——PAT甲级真题

    1022 Digital Library A Digital Library contains millions of books, stored according to their titles, ...

  8. 1022 Digital Library (30)(30 分)

    A Digital Library contains millions of books, stored according to their titles, authors, key words o ...

  9. 1022. Digital Library (30)

    A Digital Library contains millions of books, stored according to their titles, authors, key words o ...

随机推荐

  1. 【专题】计数问题(排列组合,容斥原理,Prufer序列)

    [容斥原理] 对于统计指定排列方案数的问题,一个方案是空间中的一个元素. 定义集合x是满足排列中第x个数的限定条件的方案集合,设排列长度为S,则一共S个集合. 容斥原理的本质是考虑[集合交 或 集合交 ...

  2. static_cast、dynamic_cast、reinterpret_cast、和const_c

    ;double dval = 3.14159; ival + dval;//ival被提升为double类型 2)一种类型表达式赋值给另一种类型的对象:目标类型是被赋值对象的类型 int *pi =  ...

  3. JavaScript 金额、数字、千分位、千分位、保留几位小数、舍入舍去、支持负数

    JavaScript 金额.数字 千分位格式化.保留指定位数小数.支持四舍五入.进一法.去尾法 字段说明: number:需要处理的数字: decimals:保留几位小数,默认两位,可不传: dec_ ...

  4. 内核工具 – Sparse 简介【转】

    转自:http://www.cnblogs.com/wang_yb/p/3575039.html Sparse是内核代码静态分析工具, 能够帮助我们找出代码中的隐患. 主要内容: Sparse 介绍 ...

  5. C/C++杂记:NULL与0的区别、nullptr的来历

    某些时候,我们需要将指针赋值为空指针,以防止野指针.   有人喜欢使用NULL作为空指针常量使用,例如:int* p = NULL;. 也有人直接使用0值作为空指针常量,例如:int* p = 0;. ...

  6. python面向对象(六)之元类

    元类 1. 类也是对象 在大多数编程语言中,类就是一组用来描述如何生成一个对象的代码段.在Python中这一点仍然成立: In [13]: class ObjectCreator(object): . ...

  7. 洛谷P2024食物链

    传送门啦 这道题的特殊之处在于对于任意一个并查集,只要告诉你某个节点的物种,你就可以知道所有节点对应的物种. 比如一条长为4的链 甲->乙->丙->丁 ,我们知道乙是A物种.那么甲一 ...

  8. HDU 3613 Best Reward(扩展KMP求前后缀回文串)

    题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=3613 题目大意: 大意就是将字符串s分成两部分子串,若子串是回文串则需计算价值,否则价值为0,求分割 ...

  9. Springboot + Vue + shiro 实现前后端分离、权限控制

    本文总结自实习中对项目对重构.原先项目采用Springboot+freemarker模版,开发过程中觉得前端逻辑写的实在恶心,后端Controller层还必须返回Freemarker模版的ModelA ...

  10. 为通过 ATS 检测 Tomcat 完全 TLS v1.2、完全正向加密及其结果检验

    2017 年起 app store 要求 app 对接的服务器支持 TLS v1.2,否则 ats 检测不予通过.有点强制推 TLS v1.2 的意味.本文介绍如何使 tomcat 强制执行 TLS ...