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.
Each input file contains one test case. For each case, the first line contains a positive integer N (≤104) which is the total number of books. Then N blocks follow, each contains the information of a book in 6 lines:
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:
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.
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
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
struct
数据类型表示一本书,然后放在vector里,询问的时候逐个判断。但是这个想法估计是挺耗时的。所以后来放弃了这个想法。ID
,那么我们不妨使用map<string, set<int>>
,这样查询的速度是十分快的,而且可以利用set
的插入是有序的这一点减少我们的工作量#include<bits/stdc++.h>
using namespace std;
map<string, set<int>> titles, authors, key_words, publishers, publisher_years;
void query(map<string, set<int>> &m, string pat)
{
if(m.count(pat) == 0)
cout << "Not Found" << endl;
else
{
set<int>::iterator it;
for(it=m[pat].begin();it!=m[pat].end();it++)
printf("%07d\n", *it);
}
} // 根据指定的map容器和字符串进行查找,有的话就输出
int main()
{
int books;
cin >> books;
getchar();
int id;
string tmp;
for(int i=0;i<books;i++)
{
cin >> id;
getchar();
// 分别处理书籍的各个信息,将其和对应的ID关联起来
getline(cin, tmp);
titles[tmp].insert(id);
getline(cin, tmp);
authors[tmp].insert(id);
while(cin >> tmp)
{
key_words[tmp].insert(id);
char ch = getchar();
if(ch == ‘\n‘) break;
}
getline(cin, tmp);
publishers[tmp].insert(id);
getline(cin, tmp);
publisher_years[tmp].insert(id);
}
int queries;
cin >> queries;
int num;
string pat;
for(int i=0;i<queries;i++)
{
scanf("%d: ", &num);
getline(cin, pat);
cout << num << ": " << pat << endl;
switch(num) // 根据不同类型的信息去不同的map里面查询
{
case 1 : query(titles, pat); break;
case 2 : query(authors, pat); break;
case 3 : query(key_words, pat); break;
case 4 : query(publishers, pat); break;
case 5 : query(publisher_years, pat); break;
default: break;
}
}
return 0;
}
https://pintia.cn/problem-sets/994805342720868352/problems/994805480801550336
PTA(Advanced Level)1022.Digital Library
原文:https://www.cnblogs.com/MartinLwx/p/12846808.html