PAT-1022-Digital Library

1022. Digital Library (30)

CHEN, Yue

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:

• 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

【解析】这道题其实需要用到map来进行标记，只要是属于那个下标的都要输出所以我们可以用到map<string,set<int>>.题目大意就是给你n本书的情况，然后叫你进行m次查询，然后输出该信息的书编号，如果没有这本书就输出not found比如说给你一个作者，然后会叫你输出这个作者的所有书。输入查询的时候，比如说输入1:The Testing Book意思就是输入的是书的名字，然后输出的编号就是1111111和2222222。模仿网上的代码。之前一直没思路。
#include <iostream>#include <cstdio>#include <map>#include <set>using namespace std;map<string, set<int> > title, author,publish,year,key;void facs(map< string, set<int> > &p, string &s) {    if(p.find(s) != p.end())        {        for(set<int>::iterator it = p[s].begin(); it != p[s].end(); it++)            printf("%07d\n", *it);       }    else    {       printf("Not Found\n");    }}int main() {    int i,n,m;    int id,num;    scanf("%d", &n);    string a,b,c,d,e,s;    for(int i = 0; i < n; i++) {        scanf("%d", &id);        getchar();        getline(cin,a);        title[a].insert(id);        getline(cin,b);        author[b].insert(id);//分别采取信息，编号放入其中        while(cin >>c)            {            key[c].insert(id);            char s;            s= getchar();            if(s== '\n') break;        }        getline(cin,d);        publish[d].insert(id);        getline(cin, e);        year[e].insert(id);    }    scanf("%d", &m);    for(int i = 0; i < m; i++) {        scanf("%d: ", &num);        getline(cin, s);        cout << num << ": " << s<< "\n";        if(num == 1)            facs(title, s);        else if(num == 2)        facs(author, s);        else if(num == 3)        facs(key, s);        else if(num == 4)        facs(publish,s);        else if(num ==5)        facs(year, s);    }    return 0;}