Pat(Advanced Level)Practice--1022(Digital Library)

Pat1022代码

题目描述：

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:

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

方法一：
考虑到用multimap来做，但是相同键值的元素顺序不确定，还要进行一次排序，复杂度与排序算法有关,结果最后一个case运行超时。。。

#include<cstdio>
#include<cstdlib>
#include<string>
#include<iostream>
#include<map>
#include<vector>
#include<algorithm>

using namespace std;

bool cmp(const string &l,const string &r)
{
if(l<r)
return true;
else
return false;
}

int main(int argc,char *argv[])
{
int n,m;
int i,j;
multimap<string,string> title;
multimap<string,string> author;
multimap<string,string> keyword;
multimap<string,string> publisher;
multimap<string,string> year;
scanf("%d",&n);
getchar();//吸收回车符
for(i=0;i<n;i++)
{
string num,name;
string writer,word;
string puber,time;
getline(cin,num);
getline(cin,name);
getline(cin,writer);
while(1)//截取一个单词
{
string word;
cin>>word;
keyword.insert(pair<string,string>(word,num));
if(getchar()=='\n')
break;
}
getline(cin,puber);
getline(cin,time);
title.insert(pair<string,string>(name,num));
author.insert(pair<string,string>(writer,num));
publisher.insert(pair<string,string>(puber,num));
year.insert(pair<string,string>(time,num));
}
scanf("%d",&m);
getchar();
for(i=0;i<m;i++)
{
int choice;
string query;
int flag=0;
vector<string> v;
map<string,string>::iterator it;
scanf("%d: ",&choice);
getline(cin,query);
cout<<choice<<": "<<query<<endl;
switch(choice)
{
case 1:
for(it=title.begin();it!=title.end();it++)
{
if(it->first==query)
{
flag=1;
v.push_back(it->second);
}
}
if(!flag)
cout<<"Not Found"<<endl;
break;
case 2:
for(it=author.begin();it!=author.end();it++)
{
if(it->first==query)
{
flag=1;
v.push_back(it->second);
}
}
if(!flag)
cout<<"Not Found"<<endl;
break;
case 3:
for(it=keyword.begin();it!=keyword.end();it++)
{
if(it->first==query)
{
flag=1;
v.push_back(it->second);
}
}
if(!flag)
cout<<"Not Found"<<endl;
break;
case 4:
for(it=publisher.begin();it!=publisher.end();it++)
{
if(it->first==query)
{
flag=1;
v.push_back(it->second);
}
}
if(!flag)
cout<<"Not Found"<<endl;
break;
case 5:
for(it=year.begin();it!=year.end();it++)
{
if(it->first==query)
{
flag=1;
v.push_back(it->second);
}
}
if(!flag)
cout<<"Not Found"<<endl;
break;
}
sort(v.begin(),v.end(),cmp);//键值相等的元素的位置是无序的
vector<string>::iterator iter;//所以要进行排序
for(iter=v.begin();iter!=v.end();iter++)
cout<<*iter<<endl;
}
return 0;
}

方法二：
将map与set结合起来，这样就不用进行排序了，复杂度低于方法一。
AC代码：

#include<cstdio>
#include<cstdlib>
#include<string>
#include<iostream>
#include<map>
#include<set>
#include<algorithm>

using namespace std;

int main(int argc,char *argv[])
{
int n,m;
int i,j;
map<string,set<string> > title;
map<string,set<string> > author;
map<string,set<string> > keyword;
map<string,set<string> > publisher;
map<string,set<string> > year;
map<string,set<string> >::iterator it;
scanf("%d",&n);
getchar();
for(i=0;i<n;i++)
{
string num,name;
string writer,word;
string puber,time;
getline(cin,num);
getline(cin,name);
getline(cin,writer);
while(1)
{
string word;
cin>>word;
keyword[word].insert(num);
if(getchar()=='\n')
break;
}
getline(cin,puber);
getline(cin,time);
title[name].insert(num);
author[writer].insert(num);
publisher[puber].insert(num);
year[time].insert(num);
}
scanf("%d",&m);
getchar();
for(i=0;i<m;i++)
{
int choice;
string query;
set<string>::iterator pos;
scanf("%d: ",&choice);
getline(cin,query);
cout<<choice<<": "<<query<<endl;
switch(choice)
{
case 1:
it=title.find(query);
if(it==title.end())
cout<<"Not Found"<<endl;
else
{
for(pos=it->second.begin();pos!=it->second.end();pos++)
cout<<*pos<<endl;
}
break;
case 2:
it=author.find(query);
if(it==author.end())
cout<<"Not Found"<<endl;
else
{
for(pos=it->second.begin();pos!=it->second.end();pos++)
cout<<*pos<<endl;
}
break;
case 3:
it=keyword.find(query);
if(it==keyword.end())
cout<<"Not Found"<<endl;
else
{
for(pos=it->second.begin();pos!=it->second.end();pos++)
cout<<*pos<<endl;
}
break;
case 4:
it=publisher.find(query);
if(it==publisher.end())
cout<<"Not Found"<<endl;
else
{
for(pos=it->second.begin();pos!=it->second.end();pos++)
cout<<*pos<<endl;
}
break;
case 5:
it=year.find(query);
if(it==year.end())
cout<<"Not Found"<<endl;
else
{
for(pos=it->second.begin();pos!=it->second.end();pos++)
cout<<*pos<<endl;
}
break;
}
}
return 0;
}

方法三：
最后试了一下线性搜索，居然过了，看来时间没卡的很严。。。

#include<stdio.h>
#include<string.h>
#include<stdlib.h>

typedef struct book
{
char id[10];
char title[85];
char author[85];
char keywords[100];
char publisher[85];
char year[10];
}BOOK;

BOOK book_list[10005];

int compare(const void *a,const void *b)
{
BOOK a1 = *(BOOK *)a;
BOOK b1 = *(BOOK *)b;

return strcmp(a1.id,b1.id);
}

void QueryBook(int qnum, char qword[],int nBook)
{
bool found = false;
for(int i=0;i<nBook;i++)
{
BOOK curbook = book_list[i];
switch(qnum)
{
case 1: // title
if(strcmp(qword,curbook.title)==0)
{
printf("%s\n",curbook.id);
found = true;
}
break;
case 2: // author
if(strcmp(qword,curbook.author)==0)
{
printf("%s\n",curbook.id);
found = true;
}
break;
case 3: // keyword
if(strstr(curbook.keywords,qword)!=NULL)
{
printf("%s\n",curbook.id);
found = true;
break;
}
break;
case 4: // publisher
if(strcmp(qword,curbook.publisher)==0)
{
printf("%s\n",curbook.id);
found = true;
}
break;
case 5: // year
if(strcmp(qword,curbook.year)==0)
{
printf("%s\n",curbook.id);
found = true;
}
break;
}
}
if(!found)
printf("Not Found\n");
}

int main()
{
int N,M;
scanf("%d",&N);
for(int i=0;i<N;i++)
{
scanf("%s",book_list[i].id);
getchar();
gets(book_list[i].title);
gets(book_list[i].author);
gets(book_list[i].keywords);
gets(book_list[i].publisher);
scanf("%s",book_list[i].year);
}
qsort(book_list,N,sizeof(BOOK),compare);
scanf("%d",&M);
for(int i=0;i<M;i++)
{
int qnum;
char qword[100];
scanf("%d: ",&qnum);
gets(qword);
printf("%d: %s\n",qnum,qword);
QueryBook(qnum,qword,N);
}
return 0;
}