PAT 1022 Digital Library

1022 Digital Library

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

 总结：用结构体只是写出了题目样例，还是有待进步！

有几个需要注意的点：

cin输入后在缓冲区的还存留着回车，如果此时是使用了getline(cin,s)，会读取一个回车，导致结果不正确，需要使用一个getchar()，读取回车

#include 
#include 
#include 
using namespace std;
 
struct Books{
    int id;
    string name,author;
    mapint> m;
    string publisher,year;
};
 
int main(){
    int n,k,t=0;
    cin >> n;
    Books book[n+1];
    
    for(int i=0;i
        cin >> book[i].id;
        getchar();
        getline(cin,book[i].name);
        getline(cin,book[i].author);
        char c;
        string s;
        while(1){
            cin >> s;
            book[i].m[s]=1;
            scanf("%c",&c);
            if(c=='\n') break;
        }
        getline(cin,book[i].publisher);
        getline(cin,book[i].year);
    }
    
    scanf("%d",&k);
    getchar();
    string s,w;
    for(int i=0;i
        getline(cin,s);
        t=s[0]-'0';
        set<int> v;
        if(t>=1 && t<=5)    cout << s << endl;
        w=s.substr(3);
        if(t==1){
            for(int j=0;j
                if(book[j].name==w){
                    cout << book[j].id << endl;
                }
            }
        }
        else if(t==3){
            for(int j=0;j
                if(book[j].m[w])    v.insert(book[j].id);
            }
            if(!v.size())   printf("Not Found\n");
            else for(auto it=v.begin();it!=v.end();it++) printf("%d\n",*it);
        }
        else if(t==2){
            for(int j=0;j
                if(book[j].author==w)   cout << book[j].id << endl;
            }
        }
        else if(t==4){
            for(int j=0;j
                if(book[j].publisher==w)    cout << book[j].id << endl;
            }
        }
        else if(t==5){
            for(int j=0;j
                if(book[j].year==w) cout << book[j].id << endl;
            }
        }
    }
    
    return 0;
}

还是看看到老的代码：

#include 
#include 
#include 
using namespace std;
mapint> > title, author, key, pub, year;
void query(mapint> > &m, string &str) {
    if(m.find(str) != m.end()) {
        for(auto it = m[str].begin(); it != m[str].end(); it++)
            printf("%07d\n", *it);
    } else
        cout << "Not Found\n";
}
int main() {
    int n, m, id, num;
    scanf("%d", &n);
    string ttitle, tauthor, tkey, tpub, tyear;
    for(int i = 0; i < n; i++) {
        scanf("%d\n", &id);
        getline(cin, ttitle);
        title[ttitle].insert(id);
        getline(cin, tauthor);
        author[tauthor].insert(id);
        while(cin >> tkey) {
            key[tkey].insert(id);
            char c = getchar();
            if(c == '\n') break;
        }
        getline(cin, tpub);
        pub[tpub].insert(id);
        getline(cin, tyear);
        year[tyear].insert(id);
    }
    scanf("%d", &m);
    for(int i = 0; i < m; i++) {
        scanf("%d: ", &num);
        string temp;
        getline(cin, temp);
        cout << num << ": " << temp << "\n";
        if(num == 1) query(title, temp);
        else if(num == 2) query(author, temp);
        else if(num == 3) query(key, temp);
        else if(num == 4) query(pub,temp);
        else if(num ==5) query(year, temp);
    }
    return 0;
}

 还是看看大佬的代码：

果然，大佬就是大佬，思路简单易懂

基本思路：将同一类的字符串的id都存在同一个string类型的map中，最后查找看是该字符串是否存在，存在则打印，不存在就打印Not Found

#include 
#include 
#include 
using namespace std;
mapint> > title, author, key, pub, year;
//此时用map> 这样的变量，可以再同一个字符串存储多个id,也就是只要有这个字符串的id都是可以存进去的
void query(mapint> > &m, string &str) {
    if(m.find(str) != m.end()) {
        for(auto it = m[str].begin(); it != m[str].end(); it++)
            printf("%07d\n", *it);
    } else
        cout << "Not Found\n";
}
int main() {
    int n, m, id, num;
    scanf("%d", &n);
    string ttitle, tauthor, tkey, tpub, tyear;
    for(int i = 0; i < n; i++) {
        scanf("%d\n", &id);
        getline(cin, ttitle);
        title[ttitle].insert(id);//注意这里的insert插入进去的是set，而不是map
        getline(cin, tauthor);
        author[tauthor].insert(id);
        while(cin >> tkey) {
            key[tkey].insert(id);
            char c = getchar();
            if(c == '\n') break;
        }
        getline(cin, tpub);
        pub[tpub].insert(id);
        getline(cin, tyear);
        year[tyear].insert(id);
    }
    scanf("%d", &m);
    for(int i = 0; i < m; i++) {
        scanf("%d: ", &num);
        string temp;
        getline(cin, temp);
        cout << num << ": " << temp << "\n";
        if(num == 1) query(title, temp);
        else if(num == 2) query(author, temp);
        else if(num == 3) query(key, temp);
        else if(num == 4) query(pub,temp);
        else if(num ==5) query(year, temp);
    }
    return 0;
}

好好学习，天天向上！

我要考研！