题解 |map容器解决绝大多数的查找问题

//map容器解决绝大多数的查找问题，而且由于底层是红黑树，时间复杂度较低
#include "stdio.h"
#include "string"
#include "map"
using namespace std;

int main(){
    char buf[102];
    map<string,string> myMap1;//魔咒为键
    map<string,string> myMap2;//解释为键
    while (true){
        fgets(buf,102,stdin);
        string str = buf;str.pop_back();
        if (str == "@END@")
            break;
        else{
            int pos = str.find(']');
            string magic = str.substr(1,pos-1);
            string expression = str.substr(pos+2);
            myMap1[magic] = expression;
            myMap2[expression] = magic;
        }
    }
    int m;
    scanf("%d",&m);
    getchar();
    char buf2[102];
    for (int i = 0; i < m; ++i) {
        fgets(buf2,82,stdin);
        string HaLi = buf2;HaLi.pop_back();
        if (HaLi[0] == '['){ //问的是咒语的含义，查map1
            HaLi = HaLi.substr(1,HaLi.size()-2);
            map<string,string>::iterator it = myMap1.find(HaLi);
            if (it != myMap1.end())
                printf("%s\n",it->second.c_str());
            else
                printf("what?\n");
        } else{ //问的是功能对应的咒语，查map2
            map<string,string>::iterator it = myMap2.find(HaLi);
            if (it != myMap2.end())
                printf("%s\n",it->second.c_str());
            else
                printf("what?\n");
        }
    }
}