1022. Digital Library (30)(字符串分割) 模拟

1022. Digital Library (30)

时间限制

1000 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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:

31111111The Testing BookYue Chentest code debug sort keywordsZUCS Print20113333333Another Testing BookYue Chentest code sort keywordsZUCS Print220122222222The Testing BookCYLLkeywords debug bookZUCS Print2201161: The Testing Book2: Yue Chen3: keywords4: ZUCS Print5: 20113: blablabla

Sample Output:

1: The Testing Book111111122222222: Yue Chen111111133333333: keywords1111111222222233333334: ZUCS Print11111115: 2011111111122222223: blablablaNot Found

AC代码

#include<iostream>#include<vector>#include<algorithm>#include<map>using namespace std;struct Node{    string ID;};vector<string> StrSplit(string str,char ch){    vector<string> result;    int pos = 0;    for (int i=0; i<str.size(); ++i) {        pos = (int)str.find(ch,i);        if(pos<str.size()){//找到字符串.            string  te = str.substr(i,pos-i);            result.push_back(te);            i = pos;        }else{            string  te = str.substr(i,str.size()-1);            result.push_back(te);            break;        }    }    return result;}vector<Node>allBooks;bool cmp(int a,int b){    return allBooks[a].ID<allBooks[b].ID;}int main(){    int N,M,kind;    scanf("%d%*c",&N);    map<string, vector<int>> authorMap,keyWordMap,titleMap,publisherMap,yearMap;    string title,author,keystr,publisher,year,query;    vector<string>curKeys;    allBooks.resize(N);    for(int i=0;i<N;++i)    {        getline(cin,allBooks[i].ID);        getline(cin, title);        titleMap[title].push_back(i);        getline(cin, author);        authorMap[author].push_back(i);        getline(cin, keystr);        vector<string> keys = StrSplit(keystr,' ');        for (int j=0; j<keys.size();++j)            keyWordMap[keys[j]].push_back(i);        getline(cin, publisher);        publisherMap[publisher].push_back(i);        getline(cin, year);        yearMap[year].push_back(i);    }    scanf("%d",&M);    for (int i=0; i<M;++i) {        scanf("%d: ",&kind);        getline(cin, query);        printf("%d: %s\n",kind,query.c_str());        vector<int>res;        switch (kind) {            case 1:                res.assign(titleMap[query].begin(),titleMap[query].end());                break;            case 2:                res.assign(authorMap[query].begin(),authorMap[query].end());                break;            case 3:                res.assign(keyWordMap[query].begin(),keyWordMap[query].end());                break;            case 4:                res.assign(publisherMap[query].begin(),publisherMap[query].end());                break;            case 5:                res.assign(yearMap[query].begin(),yearMap[query].end());                break;        }        if (res.size()==0)            printf("Not Found\n");        else{            sort(res.begin(),res.end(),cmp);            for (int j=0; j<res.size(); ++j)                printf("%s\n",allBooks[res[j]].ID.c_str());        }    }    return 0;}