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:

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:

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.

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: blablablaSample Output:
1: The Testing Book111111122222222: Yue Chen111111133333333: keywords1111111222222233333334: ZUCS Print11111115: 2011111111122222223: blablabla
Not Found
#include<cstdio>#include<string>#include<iostream>#include<vector>#include<map>#include<utility>using namespace std;map<int, vector<string> > mp;int main(){	int n, book_id;	scanf("%d", &n);	for(int i = 0; i < n; ++i){		vector<string> str;   //字符串向量		string temp;          //臨時字符串		scanf("%d", &book_id); getchar();		for(int j = 0; j < 5; ++j){			getline(cin, temp);			str.push_back(temp);		}		mp.insert(make_pair(book_id, str));	}	int m, query_id;	string info;	scanf("%d", &m);getchar();	for(int i = 0; i < m; ++i){		scanf("%d: ", &query_id);		getline(cin, info);		cout << query_id << ": " << info << endl;		map<int, vector<string> >::iterator it = mp.begin();		bool flag = false;		if(query_id != 3){            //查找不等于3的情況			while(it != mp.end()){				if(it->second[query_id - 1] == info){					printf("%07d/n", it->first);       //輸出書籍id					flag = true;				}				++it;			}		}else{                //查找等于3的情況			while(it != mp.end()){				if(it->second[query_id - 1].find(info) != string::npos){					printf("%07d/n", it->first);       //輸出書籍id					flag = true;				}				++it;			}		}		if(flag == false){			printf("Not Found/n");		}	}	return 0;}
這道題也可以用晴神方法，把書按查詢類型歸為一類，然后建立map<string, set<int> > 將書籍同一類id歸為一個集合例如   map<string, set<int> >  mpTitle, mpAuthor, mpKey, mpPub, mpYear;不過此題要注意對第三條關鍵字要注意，錄入的時候是整條，但查詢的時候是只查詢一個關鍵字，不是像錄入那樣整條關鍵字，所以 需要單獨拿出來， find()查找另外值得一提的是，之前用cout輸出最后一個測試點是911ms  換成printf最后一個測試點輸出時間569ms，可見輸出最好用printf，否則很容易超時