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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:
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:
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: blablablaSample 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
思路：此题看起来十分麻烦，但是需要仔细分析后思路会更加清晰。 学会利用map 和set的特性，以及二者的组合，比如说map<string,set<int> >一定要记住此处有个空格，其中访问set时需要用迭代器并且只能通过迭代器进行访问，插入时需要用insert。
1 #include<set> 2 #include<map> 3 #include<cstdio> 4 #include<algorithm> 5 #include<iostream> 6 #include<cstring> 7 #include<queue> 8 #include<vector> 9 #include<cmath> 10 using namespace std; 11 /* 12 0: id 13 1: a book title 14 2: name of an author 15 3: a key word 16 4: name of a publisher 17 5: a 4-digit number representing the year 18 */ 19 map<string,set<int> > book[6]; 20 int main(){ 21 //freopen("D:\\input.txt","r",stdin); 22 int n,m; 23 scanf("%d",&n); 24 int i,j,id; 25 string str; 26 for(i=0;i<n;i++){ 27 scanf("%d",&id); 28 getchar();//使用cin.getline()特别注意：cin.getline()是读入换行符的 29 for(j=1;j<=2;j++){ 30 getline(cin,str); 31 book[j][str].insert(id); 32 //cout<<j<<" "<<str<<endl; 33 //cout<<book[j][str].size()<<endl; 34 } 35 36 while(cin>>str){ 37 book[3][str].insert(id); 38 //cout<<3<<" "<<str<<endl; 39 //cout<<book[3][str].size()<<endl; 40 if(getchar()==‘\n‘){ 41 break; 42 } 43 } 44 45 46 for(j=4;j<=5;j++){ 47 getline(cin,str); 48 book[j][str].insert(id); 49 //cout<<j<<" "<<str<<endl; 50 //cout<<book[j][str].size()<<endl; 51 } 52 } 53 54 scanf("%d",&m); 55 for(i=0;i<m;i++){ 56 scanf("%d: ",&id); 57 printf("%d: ",id); 58 getline(cin,str); 59 cout<<str<<endl; 60 if(!book[id].count(str)){ 61 printf("Not Found\n"); 62 continue; 63 } 64 set<int>::iterator it; 65 for(it=book[id][str].begin();it!=book[id][str].end();it++){ 66 printf("%07d\n",*it);//cout<<*it<<endl; 67 } 68 } 69 return 0; 70 }
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原文地址：http://www.cnblogs.com/Deribs4/p/4765185.html
