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题目
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 ( ) which is the total number of books. Then
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
( ) which is the number of user's search queries. Then 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:
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
Sample Output:
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
思路
数据的存储与查询。
使用了最简单暴力的结构数组来存储，然后简单地按照ID排下序，遍历数组核对就行了。
这样写是很简单的，如果要使用复杂的数据结构肯定会更快。
作为参考：最后一个测试点用了660毫秒，上限为1200。
代码
最新代码@github，欢迎交流
#include
#include
#include
typedef struct book {
int ID;
char title[81];
char author[81];
char keywords[5][11];
char publisher[81];
int year;
} Book;
int cmpbyid(const void *a, const void *b)
{
return (*(Book**)a)->ID - (*(Book**)b)->ID;
}
int main()
{
int N, M, count, query_type, query_year;
char space, query_str[81];
Book books[10000] = {0}, *p, *booksbyid[10000];
scanf("%d", &N);
for(int i = 0; i < N; i++)
{
p = booksbyid[i] = books + i;
scanf("%d%c", &p->ID, &space);
scanf("%[^\n]%c", p->title, &space);
scanf("%[^\n]%c", p->author, &space);
space = '\0';
for(int k = 0; space != '\n'; k++)
scanf("%s%c", p->keywords[k], &space);
scanf("%[^\n]%c", p->publisher, &space);
scanf("%d", &p->year);
}
qsort(booksbyid, N, sizeof(Book*), cmpbyid);
scanf("%d", &M);
for(int i = 0; i < M; i ++)
{
count = 0;
scanf("%d: %[^\n]%c", &query_type, query_str, &space);
printf("%d: %s\n", query_type, query_str);
for(int j = 0; j < N; j++)
{
p = booksbyid[j];
switch(query_type)
{
case 1:
if(strcmp(query_str, p->title) == 0)
{
printf("%07d\n", p->ID);
count ++;
}
break;
case 2:
if(strcmp(query_str, p->author) == 0)
{
printf("%07d\n", p->ID);
count ++;
}
break;
case 3:
for(int k = 0; k < 5; k ++)
if(strcmp(query_str, p->keywords[k]) == 0)
{
printf("%07d\n", p->ID);
count ++;
}
break;
case 4:
if(strcmp(query_str, p->publisher) == 0)
{
printf("%07d\n", p->ID);
count ++;
}
break;
case 5:
sscanf(query_str, "%d", &query_year);
if(query_year == p->year)
{
printf("%07d\n", p->ID);
count ++;
}
break;
default:
break;
}
}
if(count == 0) puts("Not Found");
}
return 0;
}