PAT (Advanced Level) Practise 1022 Digital Library (30)
2016-03-04 23:48
381 查看
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 pat的题目基本没有卡内存和卡常数的,所以可以大胆的用stl,这题直接用map和vector就可以轻松解决了。 #include<cstdio> #include<string> #include<cstring> #include<vector> #include<iostream> #include<queue> #include<map> #include<algorithm> using namespace std; typedef long long LL; const int INF = 0x7FFFFFFF; const int maxn = 1e5 + 10; int n, tot; vector<string> p[maxn]; map<string, int> M[5]; char s[maxn], ss[maxn], sss[maxn]; int main() { scanf("%d", &n); while (n--) { scanf("%s", s); getchar(); for (int i = 0; i < 5; i++) { gets(ss); if (i == 2) { for (int k = 0, j = 0; ; k++) { if (ss[k] == ' ' || !ss[k]) { sss[j] = 0; j = 0; if (!M[i][sss]) M[i][sss] = ++tot; p[M[i][sss]].push_back(s); if (!ss[k]) break; } else sss[j++] = ss[k]; } } else { if (!M[i][ss]) M[i][ss] = ++tot; p[M[i][ss]].push_back(s); } } } scanf("%d", &n); while (n--) { scanf("%s ", s); gets(ss); int k = s[0] - '1'; printf("%s %s\n", s, ss); if (!M[k][ss]) printf("Not Found\n"); else { k = M[k][ss]; sort(p[k].begin(), p[k].end()); for (int i = 0; i < p[k].size(); i++) { cout << p[k][i] << endl; } } } return 0; }
相关文章推荐
- JavaScript原型
- Centos6.5上搭建LAMP并使用navicat工具连接数据库
- 模式的秘密——适配器模式
- java RSA签名
- Android ListView滚动条配置完全解析
- jquery追加和删除元素总结
- 工作感悟
- 对着网页进行右键操作------审查元素(快速查看标签代码)
- android设计模式之---最佳单例模式==静态内部类单例模式
- Activity详解数据传递以及隐式Intent启动
- java DOS编译相关
- Java设计模式-单例模式
- C# 数据类型的引用类型、值类型内存存储方式以及区别; 函数参数传递的引用传递(址传递)、值传递区别
- 2016第9周五
- BZOJ2040 : [2009国家集训队]拯救Protoss的故乡
- out输出参数(C#)
- 【工具】eclipse安装SVN插件
- 海量Web日志分析 用Hadoop提取KPI统计指标
- 算法_10 : 图算法_1: 图的基本概念
- linux安全第二周学习总结