字符串匹配 + 哈希表

字符串的substr()函数取子串

哈希表中count()函数判断是否存在

误区：1.没仔细看题目，原来对于给出的新饮料只能被分为两部分 2. 遍历方式错误

#include<iostream>
#include<unordered_map>
#include<cstring>
#include<vector>
#include<algorithm>
#include<cstdio>
using namespace std;

const int N = 110;
int n, m;
vector<string> drinks;
unordered_map<string, string> mp;

int main()
{
    cin >> n >> m;
    for (int i = 0; i < n; i++)
    {
        char str[N] = "";
        char cls[N] = "";
        scanf("%s%s", str, cls);
        mp[str] = cls;
        drinks.push_back(str);
    }
    for (int i = 0; i < m; i++)
    {
        string drk;
        cin >> drk;

        if (mp.count(drk)) cout << mp[drk] << endl;
        else
        {
            string rank = "";
            int count = 0;
            for (auto name : drinks)
            {
                string sub1 = drk.substr(0, name.size());//看是否为前缀
                if (sub1 == name)
                {
                    string sub2 = drk.substr(sub1.size(), drk.size());//取得剩下部分
                    if (mp.count(sub2))//看剩下部分是否已知（这个用count函数判断存在的方法之前居然没想到）
                    {
                        count++;
                        rank = mp[sub1] + mp[sub2];//这种字符串拼接方法也是第一次写，虽说跟平常写的没什么区别
                    }
                }
            }

            if (count == 1) cout << rank << endl;
            else cout << "D" << endl;
        }

    }

    return 0;
}