1022 Digital Library

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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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题目大意

模拟数字图书馆的查询功能。给出N本书的信息，以及M个需要查询的命令，command为 相应的搜索方式，command后面跟着的字符串是搜索的关键词，要求输出满⾜指定搜索书的 id。如果查询结果为NULL，输出Not Found。

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思路

map模拟即可

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C/C++ 

#include
using namespace std;
map> title,author,keys,publish,year;
void OP(map>& op,const string& key);
int main()
{
    string id,s,key;
    char ch;
    int N,command;
    cin >> N;
    getchar();
    while (N--){
        getline(cin,id);
        getline(cin,s);
        title[s].insert(id);
        getline(cin,s);
        author[s].insert(id);
        key = "";
        ch = getchar();
        while (ch!='\n'){
            if(ch==' '){
                keys[key].insert(id);
                key = "";
            }else key += ch;
            ch = getchar();
        }
        keys[key].insert(id);
        getline(cin,s);
        publish[s].insert(id);
        getline(cin,s);
        year[s].insert(id);
    }
    cin >> N;
    while (N--){
        scanf("%d: ",&command);
        getline(cin,id);
        printf("%d: %s\n",command,id.c_str());
        switch (command) {
            case 1:OP(title,id); break;
            case 2:OP(author,id);break;
            case 3:OP(keys,id);break;
            case 4:OP(publish,id);break;
            case 5:OP(year,id);break;
        }
    }
    return 0;
}
void OP(map>& op,const string& key)
{
    if(op.find(key)!=op.end()){
        for(const string& x:op[key]) cout << x << endl;
    }
    else puts("Not Found");
}

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