PAT A1022

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

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#include "stdio.h"
//#include "math.h"
#include "string.h"
#include "iostream"
//#include "stdlib.h"
#include "algorithm"
//#include "vector"
#include "set"
#include "map"
using namespace std;
//typedef long long LL;

//5个map变量分别建立书名、作者、关键词、出版社及出版年份与id的映射关系
map<string, set<int>> mpTitle, mpAuthor, mpKey, mpPub, mpYear;

void query(map<string,set<int>> &mp, string& str){//在mp中查找str
if(mp.find(str) == mp.end()) printf("Not Found\n");//找不到
else{
//找到
for (set<int>::iterator it = mp[str].begin(); it != mp[str].end(); it++) {
printf("%07d\n", *it);//输出str对应的所有id
}
}
}

int main(){
int n, m, id, type;
string title, author, key, pub, year;
scanf("%d", &n);//书的数目
for (int i = 0; i < n; i++) {
scanf("%d%*c", &id);
getline(cin, title);//书名
mpTitle[title].insert(id);//把id加入title对应的集合中
getline(cin, author);//读入
mpAuthor[author].insert(id);//把id加入author对应的集合中
while (cin >> key) {//每次读入一个关键词key
mpKey[key].insert(id);//把id加入key对应的集合中
int c = getchar();//接受关键字key后的字符
if (c == '\n') {//如果是换行,说明关键字输入结束
break;
}
}
getline(cin, pub);//出版社pub
mpPub[pub].insert(id);//把id加入pub对应的集合中
getline(cin,year);//出版日期year
mpYear[year].insert(id);//把id加入year对应的集合中
}
string temp;
scanf("%d", &m);//查询次数
for (int i = 0 ; i < m; i++) {
scanf("%d: ", &type);//查询类型
getline(cin,temp);//欲查询的字符串
cout << type << ": " << temp << endl;//输出类型和字符串
if (type == 1) {
query(mpTitle, temp);
}else if (type == 2) {
query(mpAuthor, temp);
}else if (type == 3) {
query(mpKey, temp);
}else if (type == 4) {
query(mpPub, temp);
}else{
query(mpYear, temp);
}
}
return 0;
}

介绍map<string,set>用法:
set元素插入采用的是insert函数,因此mp[“Name”].insert(id);
遍历问题:

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for(set<int>::iterator it = mp["Mike"].begin(); it != mp["Mike"].end() ; it++){
printf("%d\n", *it);
}

其他的如书名,作者,出版社及出版日期可以作为整体读入
关键词读入的问题:

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while(cin >> key){
mpKey[key].insert(id);
char c = getchar();//读入一个词后的字符
if(c == '\n') break;//如果是换行,跳出
}