Digital Library（c/c++）

题目要求：

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.

输入要求：

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

输出要求：

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.

输入样例：

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

输出样例：

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

这题吧，逻辑倒不是很难，就是找到好方法。
熟悉掌握vector数组，scanf，printf，以及字符串的一些读取，分割。
↓这一块是新学到的一些东西，读取关键字时有些麻烦，先用key接收读取的一行关键字，然后用istringstream函数（包含头文件#include < sstream >），将一行关键字进行分割。存入到book[i].key.push_back(temp)中。

		string key;
		getline(cin,key);
		istringstream cinKeyWord(key); //用key作为输入流输入每一个关键词
		string temp;
	    while(cinKeyWord >> temp){
     
			book[i].key.push_back(temp);
	    }

对id进行排序。
最后按类别进行筛选。
代码如下：

#include
#include
#include
#include
#include
#include 
using namespace std;

struct books{
     
	int id;
	string title;//书名 
	string author;
	vector<string>key;
	string publisher;
	string year;
};

bool com(books a,books b){
     
	return a.id<b.id;
}

int main(){
     
	int N;cin>>N;
	vector<books> book(N);
	for(int i=0;i<N;i++){
     
		scanf("%d\n",&book[i].id);
		getline(cin,book[i].title);
		getline(cin,book[i].author);
		string key;
		getline(cin,key);
		istringstream cinKeyWord(key); //用key作为输入流输入每一个关键词
		string temp;
	    while(cinKeyWord >> temp){
     
			book[i].key.push_back(temp);
	    }
		getline(cin,book[i].publisher);
		getline(cin,book[i].year);
	}
	sort(book.begin(),book.end(),com);
	int M;cin>>M;
	for(int j=0;j<M;j++){
     
		int flag,cnt = 0;
		string info;
		scanf("%d: ",&flag);
		getline(cin,info);
		printf("%d: %s\n",flag,info.c_str());
	    if(flag == 1){
     
	    	for(int i=0;i<N;i++){
     
	    		if(book[i].title == info){
     
	    			printf("%07d\n",book[i].id);cnt++;
				}
			}
		}else if(flag == 2){
     
			for(int i=0;i<N;i++){
     
				if(book[i].author == info){
     
					printf("%07d\n",book[i].id);cnt++;
				}
			}
		}else if(flag == 3){
     
			for(int i=0;i<N;i++){
     
				for(int k=0;k<book[i].key.size();k++){
     
					if(book[i].key[k] == info){
     
						printf("%07d\n",book[i].id);cnt++;
					}
				}
			}
		}else if(flag == 4){
     
			for(int i=0;i<N;i++){
     
				if(book[i].publisher == info){
     
					printf("%07d\n",book[i].id);cnt++;
				}
			}
		}else if(flag == 5){
     	
			for(int i=0;i<N;i++){
     
				if(book[i].year == info){
     
					printf("%07d\n",book[i].id);cnt++;
				}
			}
		}
		if(cnt == 0){
     
			printf("Not Found\n");
		}
	}
	return 0;
}

当然，也有更为简洁的代码，但本人能力有限55555
上学时间又回来了，满满的期待！

一个集坚强与自信于一身的菇凉。