pat 甲级 1022 Digital Library

啥都别说了
智障了

看着输入输出觉得很复杂
其实解决了怎么存储，剩下的比较起来很简单
循环就可以了
没用别人那种精巧的方法
直接建结构体存储，没用到map

知识点： vector用法； getline；
C语言没办法对string进行输入输出，用cin cout比较好

写完了一直不过，因为自己忘了输出要查询的信息

#include <cstdio>
#include <iostream>
#include <map>
#include <vector>
#include <algorithm>

using namespace std;

int n, m;


struct bo{

    string id;
    string title;
    string author;
    vector<string> keyword;
    string publisher;
    string year;
};

map<string,vector<string> > keymap;
bo book[11000];

bool cmp(bo a, bo b){
    return a.id < b.id;
}

int main(){

    scanf("%d",&n); getchar();
    for(int i=0; i<n; i++){

        string id, title, author, publisher, key, year;
        getline(cin,id); book[i].id = id; 
        getline(cin,title); book[i].title = title; 
        getline(cin,author); book[i].author = author;
        while(cin >> key){
            book[i].keyword.push_back(key);
            char c = getchar();
            if(c == '\n') break;
        }
        getline(cin,publisher); book[i].publisher = publisher;
        getline(cin,year); book[i].year = year;

    }

    sort(&book[0],&book[n],cmp);
    scanf("%d",&m);
    for(int j=0; j<m; j++){
        int index;
        int flag = 0;
        string in;
        scanf("%d: ",&index);
        getline(cin,in);
        cout << index << ": " << in << endl;        
        switch(index){

            case 1: for(int k=0; k<n; k++){
                      if(book[k].title == in) {
                        cout << book[k].id << endl;
                        flag = 1;
                      }                                 
                    }
                    if(flag == 0) cout << "Not Found" << endl;
                    break;

            case 2: for(int k=0; k<n; k++){
                      if(book[k].author == in) {
                        cout << book[k].id << endl;
                        flag = 1;
                      }                                 
                    }
                    if(flag == 0) cout << "Not Found" << endl;
                    break;

            case 3: for(int k=0; k<n; k++){
                        for(int j=0; j<book[k].keyword.size(); j++){
                            if(book[k].keyword[j] == in) {
                            cout << book[k].id << endl;
                            flag = 1;
                            }       
                        }                                           
                    }
                    if(flag == 0) cout << "Not Found" << endl;
                    break;

            case 4: for(int k=0; k<n; k++){
                      if(book[k].publisher == in) {
                        cout << book[k].id << endl;
                        flag = 1;
                      }                                 
                    }
                    if(flag == 0) cout << "Not Found" << endl;
                    break;

            case 5: for(int k=0; k<n; k++){
                      if(book[k].year == in) {
                        cout << book[k].id << endl;
                        flag = 1;
                      }                                 
                    }
                    if(flag == 0) cout << "Not Found" << endl;
                    break;
        }
    }



    return 0;
}

啊！