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.
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 Nblocks 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 title2: name of an author3: a key word4: name of a publisher5: a 4-digit number representing the yearFor 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.
没写完，又臭又长！key word处理方法没写出来，参考柳神代码：
#include <iostream> #include <string> #include <cstring> #include <vector> #include <cstdlib> #include <algorithm> using namespace std; int n, m; struct book { int id; char title[80]; char author[80]; char key[60]; char publisher[80]; int pub_year; }books[10010]; bool cmp(int a, int b) { return a < b; } vector<int> title_q; vector<int> author_q; vector<int> key_q; vector<int> publisher_q; vector<int> pub_year_q; void look_for_titile(char t[]) { for(int i = 0; i < n; i++) { if(strcmp(t, books[i].title) == 0) { title_q.push_back(books[i].id); } } sort(title_q.begin(),title_q.end(),cmp); if (title_q.size() == 0) { printf("Not Found\n"); }else { for (int i = 0; i < title_q.size(); i++) { printf("%d\n",title_q[i]); } } } void look_for_author(char t[]) { for(int i = 0; i < n; i++) { if(strcmp(t, books[i].author) == 0) { author_q.push_back(books[i].id); } } sort(author_q.begin(),author_q.end(),cmp); if (author_q.size() == 0) { printf("Not Found\n"); }else { for(int i = 0; i < author_q.size(); i++) { printf("%d\n",author_q[i]); } } } void look_for_key(char t[]) { char keys[5], ind = 0; for(int i = 0; i < strlen(t); i++) { string s = ""; if(t[i] != ' ') { s += t[i]; }else { keys[ind] = s; s = ""; } } } void look_for_publisher(char t[]) { for(int i = 0; i < n; i++) { if(strcmp(t, books[i].publisher) == 0) { publisher_q.push_back(books[i].id); } } sort(publisher_q.begin(),publisher_q.end(),cmp); if (publisher_q.size() == 0) { printf("Not Found\n"); }else { for (int i = 0; i < publisher_q.size(); i++) { printf("%d\n",publisher_q[i]); } } } void look_for_pub_year(int t) { for(int i = 0; i < n; i++) { if(t == books[i].pub_year) { pub_year_q.push_back(books[i].id); } } sort(pub_year_q.begin(),pub_year_q.end(),cmp); if (pub_year_q.size() == 0) { printf("Not Found\n"); }else { for (int i = 0; i < pub_year_q.size(); i++) { printf("%d\n",pub_year_q[i]); } } } int main() { scanf("%d", &n); for(int i = 0; i < n; i++) { scanf("%d", &books[i].id); getchar(); cin.getline(books[i].title,80); cin.getline(books[i].author,80); cin.getline(books[i].key,60); cin.getline(books[i].publisher,80); scanf("%d", &books[i].pub_year); } scanf("%d", &m); getchar(); while(m--) { char query[80]; cin.getline(query,80); printf("%s\n", query); char num = query[0]; char* q = query+3; switch(num) { case '1': { look_for_titile(q); break; } case '2': { look_for_author(q); break; } case '3': { look_for_key(q); break; } case '4': { look_for_publisher(q); break; } case '5': { look_for_pub_year(atoi(q)); break; } } } return 0; }柳神的代码：
收获：看大神的代码真的太太太震撼了！！
1.map映射，把一个string映射成一个set，还自动去重！太棒了叭！
2.还有因为不确定key word，采用while(cin >> s)并且判断c = getchar()，c是否等于\n退出循环的方式也叫绝...
3.getline（）前面有换行符要用getchar（）接收，或者 采用scanf("%d\n",&id)避免掉getchar();
何时我才能有这样的思维！！
#include <iostream> #include <map> #include <set> #include <string> using namespace std; map<string, set<int> > title, author, key, pub, year; void query(map<string, set<int> > &m, string &str) { if(m.find(str) != m.end()) { for(auto it = m[str].begin(); it != m[str].end(); it++) printf("d\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; }
