PAT (Advanced Level) Practise 1022 Digital Library (30)

1022. Digital Library (30)

时间限制

1000 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

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: blablabla
Not Found
pat的题目基本没有卡内存和卡常数的，所以可以大胆的用stl，这题直接用map和vector就可以轻松解决了。
#include<cstdio>#include<string>#include<cstring>#include<vector>#include<iostream>#include<queue>#include<map>#include<algorithm>using namespace std;typedef long long LL;const int INF = 0x7FFFFFFF;const int maxn = 1e5 + 10;int n, tot;vector<string> p[maxn];map<string, int> M[5];char s[maxn], ss[maxn], sss[maxn];int main(){scanf("%d", &n);while (n--){scanf("%s", s); getchar();for (int i = 0; i < 5; i++){ gets(ss); if (i == 2) { for (int k = 0, j = 0; ; k++) { if (ss[k] == ' ' || !ss[k]) { sss[j] = 0;j = 0; if (!M[i][sss]) M[i][sss] = ++tot; p[M[i][sss]].push_back(s); if (!ss[k]) break; } else sss[j++] = ss[k]; } } else  { if (!M[i][ss]) M[i][ss] = ++tot; p[M[i][ss]].push_back(s); }}}scanf("%d", &n);while (n--){scanf("%s ", s); gets(ss);int k = s[0] - '1';printf("%s %s\n", s, ss);if (!M[k][ss]) printf("Not Found\n");else{k = M[k][ss];sort(p[k].begin(), p[k].end());for (int i = 0; i < p[k].size(); i++){cout << p[k][i] << endl;}}}return 0;}