【奇淫技巧】C++理解继承+结构体封装 好题 Gym - 101673E Is-A? Has-A? Who Knowz-A?

题目链接
比赛链接
题意：给定n个表达式，m个判断，每次有A is B,A has B
会有 A is B,B is C = A is C
A has B,b has c ,A has C
A has B ,B is C ,A has C
A is B,B has C, A has C
这样的合并，请你输出判断的结果。

#include <iostream>
#include<bits/stdc++.h>
using namespace std;
struct know :map<string,int>
{
    int operator ()(const string &tmp)
    {
        if(!count(tmp))
        {
            insert(make_pair(tmp,size()));
        }
        return at(tmp);
    }
}id;
struct node
{

    struct Vex
    {
        vector<int>a;
        int s[2][512];
    };
    struct road
    {
       int from,to,op;
    };
    int fa;
    vector<Vex>v;
    vector<road>e;
    node(int n):v(n){}
    void add(const road &st)
    {
       v[st.from].a.push_back(e.size());
       //这里用到了类似前向星的东西
       e.push_back(st);

    }
    void dfs(int u,int is)
    {
       v[fa].s[is][u] = 1;
       int to ;
//       int len =v[u].a.size();
//       for(int i=0;i<len;i++)
//       {
//         int k = v[u].a[i];
//         to = e[k].to;
//         int yy = k;
//         k = is&&e[yy].op;
//         if(!v[fa].s[k][to])
//         {
//             dfs(to,fa,k);
//         }
//       }
       for(int y:v[u].a)
       {
           //
           to = e[y].to ;
           int jj = is&&e[y].op;
          if(!v[fa].s[jj][to])
          {
              dfs(to,jj);
          }
       }
    }
}exm(512);//注意这里需要对vector的大小进行初始化
int n,m;
int main()
{

    cin>>n>>m;
    string t1,t2,t3;
    node::road tmp;
    for(int i=1; i<=n; i++)
    {
        cin>>t1>>t2>>t3;
        tmp.from = id(t1);
        tmp.to  =id(t3);
        tmp.op  = (t2=="is-a");
        exm.add(tmp);
    }
   for(int i=0;i<(int )id.size();i++)
   {
      exm.fa = i;
      exm.dfs(i,1);
   }
    for(int i=1; i<=m; i++)
    {
       cin>>t1>>t2>>t3;
       int j =  (t2=="is-a"?1:0);
       int f= exm.v[id(t1)].s[t2=="is-a"][id(t3)];
       if(f)printf("Query %d: true\n",i);
       else printf("Query %d: false\n",i);
    }

    return 0;
}