(PAT) 1022 Digital Library (利用映射map存储数据,以及输入输出控制)
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 (≤104) 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
解题思路:
我们使用 map
使用Map[“Mike”].insert(1001) 就可以实现加入
接下来是读入数据的问题
使用cin读取字符串的话,空格会被当成结束符,所以如果输入The Testing book的话,cin只会读入The
要读入空格的话,必须使用getline(cin,string)
同时,在用scanf()读取数据后,必须用getchar()将换行吸收掉,否则getline会把换行读入
关于这题,创建5个map存储对应的ID,然后根据用户的输入数据查询即可
#include<iostream>#include<algorithm>#include<map>#include<string.h>#include<string>#include<set>using namespace std;map<string, set<int>> titleMap, authorMap, keyWordsMap, publisherMap, publisherYearMap; //存5张mapvoid searchInfo(map<string, set<int>>& MAP, string& content) {if (MAP.find(content) == MAP.end()) {cout << "Not Found" << endl;}else {for (auto itr = MAP[content].begin(); itr != MAP[content].end(); ++itr) {printf("%07d\n", *itr);}}}int main() {int N, M;char c; //用来吸收掉空格scanf("%d", &N);for (int i = 0; i < N; ++i) {string input;int ID;scanf("%d", &ID);c = getchar();getline(cin, input);titleMap[input].insert(ID);getline(cin, input);authorMap[input].insert(ID);while (cin >> input) {keyWordsMap[input].insert(ID);c = getchar();if(c == '\n') break; //如果是换行的话,退出}getline(cin, input);publisherMap[input].insert(ID);getline(cin, input);publisherYearMap[input].insert(ID);}scanf("%d", &M);for (int i = 0; i < M; ++i) {int queryId;string queryContent;scanf("%d: ", &queryId);getline(cin, queryContent);cout << queryId << ": " << queryContent << endl;switch (queryId){case 1:searchInfo(titleMap, queryContent);break;case 2:searchInfo(authorMap, queryContent);break;case 3:searchInfo(keyWordsMap, queryContent);break;case 4:searchInfo(publisherMap, queryContent);break;case 5:searchInfo(publisherYearMap, queryContent);break;default:break;}}system("PAUSE");return 0;}
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