1022. Digital Library (30)
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1022. Digital Library (30)
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:
31111111The Testing BookYue Chentest code debug sort keywordsZUCS Print20113333333Another Testing BookYue Chentest code sort keywordsZUCS Print220122222222The Testing BookCYLLkeywords debug bookZUCS Print2201161: The Testing Book2: Yue Chen3: keywords4: ZUCS Print5: 20113: blablabla
Sample Output:
1: The Testing Book111111122222222: Yue Chen111111133333333: keywords1111111222222233333334: ZUCS Print11111115: 2011111111122222223: blablablaNot Found
#include<cstdio>#include<iostream>#include<string>#include<algorithm>#include<vector>#include<map>using namespace std;typedef int book;int ID[10001];map<string,vector<book> > mf;//核心数据结构,一个关键词对应一个book指针数组 void FindAnswer(string s){ if(mf.find(s)!=mf.end()){ sort(mf[s].begin(),mf[s].end());//我选择最后排序 for(int i=0;i!=mf[s].size();i++) printf("%07d\n",mf[s][i]); } else cout<<"Not Found"<<endl;}void BuildMap(string s,int i){ vector<book> empty; if(mf.find(s)==mf.end()) mf.insert(pair<string,vector<book> >(s,empty)); mf[s].push_back(ID[i]);}int main(){ int N,i,j,M; string t,search; cin>>N; for(i=0;i<N;i++){//输入并建立哈希表 cin>>ID[i]; cin.get(); for(j=0;j<5;j++){ char c=' '; string s; if(j!=2){//keyword部分需要单独考虑,其他的部分其实都是一样的 getline(cin,s); BuildMap(s,i); } else{ while(c!='\n'){ c=cin.get(); if(c!='\n'&&c!=' ') s+=c; else if(c==' '){ BuildMap(s,i); s.clear(); } } BuildMap(s,i); } } } cin>>M; while(M--){ cin>>t; cin.get();//去除上一个cin留下的回车符号 cout<<t<<" "; getline(cin,search); cout<<search<<endl; FindAnswer(search); search.clear(); }}
- 1022. Digital Library (30)
- 1022. Digital Library (30)
- 1022. Digital Library (30)
- 1022. Digital Library (30)
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- 1022. Digital Library (30)
- 1022. Digital Library (30)
- 1022. Digital Library (30)
- 1022. Digital Library (30)
- 1022. Digital Library (30)
- 1022. Digital Library (30)
- 1022. Digital Library (30)
- 1022. Digital Library (30)
- 1022. Digital Library (30)
- 1022. Digital Library (30)
- 1022. Digital Library (30)
- 1022. Digital Library (30)
- 1022. Digital Library (30)
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