题目如下：
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
这道题目我参考了sunbaigui的解法，这是一道通过属性值来找记录的问题，属于倒排索引，由于属性值较多，需要使用多个map进行存储，对于多个ID公用多个属性值的问题，可以把map的ID那一维设置为vector，从而可以容纳多个ID，为了满足ID的升序输出，需要对每个map中的记录按照ID升序进行排序。
通过学习sunbaigui的代码，我学到了一些细节如下：
1.map可以通过索引值直接插入：
例如map<string,int> mm 一般的插入方式为mm.insert(pair<string,int>("str",100))
还可以通过mm[“str”] = 100来实现插入
2.由于题目中的字符串有空格出现，因此应该使用getline(cin,str)来获取每一个字符串，注意使用getline时如果前面有其他类型的输入，例如cin和scanf，应当加一个getchar()吃掉回车符。
3.对于一个以空格分隔的多个关键词组成的字符串，要提取出每一个部分，使用sstream头文件中的istringstream来分离每个部分，设keywords中存储着多个以空格分隔的关键词，具体实现为：
istringstream istr(keywords);
while(!istr.eof()){
string keyword;
istr >> keyword;
// 此时keyword中存的为一个关键词，istr每输出一次就后移一个，直到EOF
}
4.map的find函数只能找第一维的内容。
5.要对容器排序，首先保证容器内存储的类型有<符，然后调用sort函数传入begin和end迭代器。
题目的具体实现为：
定义5个map，每个map的第一维为string，第二维为vector<string>，其中第一维保存不同的属性值，第二维保存各个属性值对应的ID，在输入记录的过程中不断把记录存入map，接着对第二维进行排序，这时候得到的所有记录就是按照ID的升序排列的了，在查找时对不同的查找类型选择不同的map，如果找到，则可以得到一个ID容器，输出容器中所有ID即可，找不到则输出Not Found。
由于输入缓冲区和输出缓冲区是分离的，因此可以在输入一条记录后立即打印一条结果。
#include<iostream>
#include<string>
#include<vector>
#include<algorithm>
#include<map>
#include<sstream>
#include<stdio.h>
using namespace std;
int main()
{
map<string,vector<string> > infoMaps[5];
string ID,title,author,keywords,publisher,year;
int N;
scanf("%d",&N);
for(int i = 0; i < N; i++){
getchar(); // 吃掉每次输入结尾的回车。
getline(cin,ID);
getline(cin,title);
getline(cin,author);
getline(cin,keywords);
getline(cin,publisher);
cin >> year;
infoMaps[0][title].push_back(ID);
infoMaps[1][author].push_back(ID);
infoMaps[3][publisher].push_back(ID);
infoMaps[4][year].push_back(ID);
istringstream istr(keywords);
while(!istr.eof()){
string keyword;
istr >> keyword;
infoMaps[2][keyword].push_back(ID);
}
}
for(int i = 0; i < 5; i++){
map<string, vector<string> >::iterator it;
for(it = infoMaps[i].begin(); it!=infoMaps[i].end(); it++){
sort(it->second.begin(),it->second.end());
}
}
cin >> N;
int index;
string query;
for(int i = 0; i < N; i++){
scanf("%d: ",&index);
getline(cin,query);
cout << index << ": " << query << endl;
map<string, vector<string> >::iterator it;
it = infoMaps[index - 1].find(query);
if(it != infoMaps[index - 1].end()){
vector<string> IDs = it->second;
for(int cnt = 0; cnt < IDs.size(); cnt++){
cout << IDs[cnt] << endl;
}
}else{
cout << "Not Found" << endl;
}
}
return 0;
}