1022 Digital Library (30 point(s))
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:
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这道题难度不大，但是比较烦！
最开始看到限定关键词和出版商的数量，我还以为会有样例卡这个，本来想建一个map<string,vector<int> >来在输入的时候存储关键词和出版商id的，实际证明不需要。
注意点：
1. 如果要在cin或者scanf后紧接着用getline()，需要用cin.get()或者getchar()先吸收回车符；
参考他人代码的时候看到不少有趣的方法：
1. 拆解单词的方法
a. 我的方法。读取后再拆解，注意到最后若word不为空，则这也是一个单词。
getline(cin,s);
string word="";
for(int j=0;j<s.size();j++){
if(s[j]!=' '){
word+=s[j];
}
else if(s[j]==' '&&!word.empty()){
B[i].keywords.push_back(word);
word = "";
}
}
if(!word.empty()) B[i].keywords.push_back(word);b. 更简单的方法，简直是处理字符串的神器。
while(cin>>s){
B[i].keywords.push_back(s);
char ch = getchar();
if(ch=='\n') break;
}2. 不必要去分解单词，再查询时去查询子串即可。
参考链接：https://blog.youkuaiyun.com/eric4784510/article/details/81664680
vector<string> searchKeyword(string keyword){
vector<string> ans;
for(int i=0;i<n;i++){
if(b[i].keywords.find(keyword)!=string::npos)
ans.push_back(b[i].id);
}
return ans;
}3. 建立map<string, set<int>> mpTitle, mpAuthor, mpKey, mpPub, mpYear的映射关系，分别记录每个标题、作者、关键词、出版商、年份对应的id，又利用set的排序自动完成了对id的排序
参考链接：https://blog.youkuaiyun.com/qq_24452475/article/details/99851018
#include<map>
#include<set>
#include<string>
#include<iostream>
using namespace std;
void query(map<string, set<int>>& mp, string& str)
{
if(mp.find(str) == mp.end())
printf("Not Found\n");
else{
for(set<int>::iterator it=mp[str].begin(); it!=mp[str].end(); it++)
printf("%07d\n", *it);
}
}
int main()
{
int n, m, id, type;
scanf("%d", &n);
string title, author, key, pub, year;
map<string, set<int>> mpTitle, mpAuthor, mpKey, mpPub, mpYear;
for(int i=0; i<n; i++)
{
scanf("%d", &id);
char c = getchar();
getline(cin, title);
mpTitle[title].insert(id);
getline(cin, author);
mpAuthor[author].insert(id);
while(cin >> key)
{
mpKey[key].insert(id);
c = getchar();
if(c == '\n') break;
}
getline(cin, pub);
mpPub[pub].insert(id);
getline(cin, year);
mpYear[year].insert(id);
}
string searchKey;
scanf("%d", &m);
for(int i=0; i<m; i++)
{
scanf("%d: ", &type);
getline(cin, searchKey);
cout << type << ": " << searchKey << endl;
if(type == 1) query(mpTitle, searchKey);
else if(type==2) query(mpAuthor, searchKey);
else if(type==3) query(mpKey, searchKey);
else if(type==4) query(mpPub, searchKey);
else query(mpYear, searchKey);
}
return 0;
}
我的AC代码
#include<iostream>
#include<vector>
#include<algorithm>
#include<map>
#include<cstring>
using namespace std;
int N,M;
struct book{
int id;
string title,author,publisher,year;
vector<string> keywords;
bool operator < (const book &other) const{
return id<other.id;
}
};
book B[10007];
void query(char c,string s){
bool found = false;
if(c=='1'){
for(int i=0;i<N;i++){
if(B[i].title==s){
printf("%07d\n",B[i].id);
found = true;
}
}
}
else if(c=='2'){
for(int i=0;i<N;i++){
if(B[i].author==s){
printf("%07d\n",B[i].id);
found = true;
}
}
}
else if(c=='3'){
for(int i=0;i<N;i++){
for(int j=0;j<B[i].keywords.size();j++){
if(B[i].keywords[j]==s){
printf("%07d\n",B[i].id);
found = true;
break;
}
}
}
}
else if(c=='4'){
for(int i=0;i<N;i++){
if(B[i].publisher==s){
printf("%07d\n",B[i].id);
found = true;
}
}
}
else if(c=='5'){
for(int i=0;i<N;i++){
if(B[i].year==s){
printf("%07d\n",B[i].id);
found = true;
}
}
}
if(!found) cout<<"Not Found"<<endl;
}
int main(void){
cin>>N;
string s;
for(int i=0;i<N;i++){
cin>>B[i].id;getchar();
getline(cin,s);
B[i].title=s;
getline(cin,s);
B[i].author=s;
getline(cin,s);
string word="";
for(int j=0;j<s.size();j++){
if(s[j]!=' '){
word+=s[j];
}
else if(s[j]==' '&&!word.empty()){
B[i].keywords.push_back(word);
word = "";
}
}
if(!word.empty()) B[i].keywords.push_back(word);
getline(cin,s);
B[i].publisher=s;
getline(cin,s);
B[i].year=s;
}
sort(B,B+N);
cin>>M;getchar();
char c;
while(M--){
cin>>c;cin.get();cin.get();
getline(cin,s);
cout<<c<<": "<<s<<endl;
query(c,s);
}
return 0;
}
261
扫一扫
被折叠的 条评论
为什么被折叠?
到【灌水乐园】发言
