1022 Digital Library

1022 Digital Library

题目大意

输入书籍信息，然后根据顾客查找的信息，返回对应书籍的id，有可能不止一个id

算法思想

• 用map会简单很多，而且map会将关键字按大小排列，刚好符合题目要求
• 输入有些困难，不过以后熟悉了就好，因为要接收带空格的字符串
• 别忘了每次cin后要跟着一个输入多余回车的函数
• 题目较复杂，但是不算难题，仔细分析就能求出
• 没啥特殊测试点，写出来就是满分

代码

#include<iostream>
#include<map>
#include<string>
#include<vector>
#include<algorithm>
using namespace std;
struct book//书籍内容
{
	string title, author, publisher, year;
	vector<string>keywords;
};
struct query//询问内容
{
	int wb;
	string con;
};
map<string, book>library;
vector<query>cust;
void find_b(int x, string s)//查找字典，输出对应的id
{
	bool flag = false;
	auto it = library.begin();
	if (x == 1)
		for (; it != library.end(); it++)
		{
			if (s == it->second.title)
			{
				flag = true;
				cout << it->first << endl;
			}
		}
	else if (x == 2)
		for (; it != library.end(); it++)
		{
			if (s == it->second.author)
			{
				flag = true;
				cout << it->first << endl;
			}
		}
	else if (x == 3)
		for (; it != library.end(); it++)
		{
			if (find(it->second.keywords.begin(), it->second.keywords.end(), s) != it->second.keywords.end())//vector内查找
			{
				flag = true;
				cout << it->first << endl;
			}
		}
	else if (x == 4)
		for (; it != library.end(); it++)
		{
			if (s == it->second.publisher)
			{
				flag = true;
				cout << it->first << endl;
			}
		}
	else if (x == 5)
		for (; it != library.end(); it++)
		{
			if (s == it->second.year)
			{
				flag = true;
				cout << it->first << endl;
			}
		}
	if (!flag)
		cout << "Not Found" << endl;
	return;
}
int main()
{
	int n=0, m;
	cin >> n;
	cin.get();//吃掉回车
	for (int i = 0; i < n; i++)//输入书籍信息
	{
		string id,word;
		getline(cin, id);
		getline(cin, library[id].title);
		getline(cin, library[id].author);
		while (cin >> word)//输入每个关键字
		{
			library[id].keywords.push_back(word);
			if (cin.get() == '\n') break;
		}
		getline(cin, library[id].publisher);
		cin >> library[id].year;
		cin.get();
	}
	cin >> m;
	cin.get();
	cust.resize(m);
	for (int i = 0; i < m; i++)//输入询问内容
	{
		int num;
		string s;
		cin >> num;
		getchar();
		getchar();
		getline(cin, s);
		cust[i].wb = num;
		cust[i].con = s;
	}
	for (int i = 0; i < m; i++)//输出内容
	{
		cout << cust[i].wb << ": " << cust[i].con << endl;
		find_b(cust[i].wb,cust[i].con);
	}
	return 0;
}