PAT甲级1022

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最新推荐文章于 2025-02-09 13:55:00 发布

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分类专栏： PAT 文章标签：算法

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本文针对PAT甲级1022题“数字图书馆”进行详细解析，介绍了如何根据读者查询条件检索并排序图书ID的方法。该题涉及图书信息存储、字符串匹配及排序算法等内容。

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PAT甲级1022

原题：
1022 Digital Library (30分)
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 (≤10
4
) 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

题目大意：给出一些图书的信息，然后通过某个标签去查找图书的编号

注意点：
1、除了第4个关键词以外，其余信息全是完全匹配，关键词这个里面只要包括带查找的字符串就OK了，不需要完全一样
2、不要图方便省略if-else大括号(这个楼主刚试完毒)
3、查询的时候边输入边处理
4、凡是用到getline一定要记得清缓存(在scanf/cin之后用cin.ignore()或者fflush(这个以前用过现在用会报错，未解决)或者getchar())

#include <iostream>
#include <string>
#include <vector>
#include <algorithm>
using namespace std;

struct book{	//记录书籍的信息
	vector<string> info;	//0：ID，1：title，2：author，3：keywords，4：publisher，5：publishyear
};
struct query{	//记录查询信息
	int index;	//查询编号
	string content;	//查询内容
};
bool cmp(book a,book b){
	return a.info[0]<b.info[0];
}
int main(){
	int n,m;
	scanf("%d",&n);
	cin.ignore();
	book b[n];
	for(int i=0;i<n;i++)
		for(int j=0;j<6;j++){
			string tmp;
			getline(cin,tmp);
			b[i].info.push_back(tmp);
		}
	sort(b,b+n,cmp);
	scanf("%d",&m);
	cin.ignore();
	query q[m];
	for(int i=0;i<m;i++){
		scanf("%d:",&q[i].index);
		cin.ignore();
		getline(cin,q[i].content);
		printf("%d: ",q[i].index);
		cout<<q[i].content<<endl;
		bool flag=false;	//记录是否有找到
		for(int j=0;j<n;j++){
			if(q[i].index!=3){	//若查找的不是关键词
				if(b[j].info[q[i].index]==q[i].content) cout<<b[j].info[0]<<endl,flag=true;
			}
			else{
				if(b[j].info[3].find(q[i].content)!=string::npos) cout<<b[j].info[0]<<endl,flag=true;
			}
		}
		if(flag==false) printf("Not Found\n");
	}
	system("pause");
	return 0;
}