How to solve manager peer problem question?

I was asked below question in google interview but could not answer it. I did google search but could not find any correct solution for this problem. Can you please help me to provide a solution for this problem?

Design a data structure to support following functions:

1. setManager(A, B) sets A as a direct manager of B
2. setPeer(A, B) sets A as a colleague of B. After that, A and B will have the same direct Manager.
3. query(A, B) returns if A is in the management chain of B. Every person only has 1 direct manager.

Use three hash tables 1. employee to group id mapping (emp) 2. group id to list of employees (gid) 3. group id -> manager id table (manager)

int id = 0;
void setpeer(string A, string B) {
int id1 = 0;
int id2 = 0;
if (emp.find(A) != emp.end())
   id1 = emp[A];
if (emp.find(B) != emp.end())
   id2 = emp[B];

if (!id1 && !id2) {
    ++id;
    emp[A] = id;
    emp[B} = id;
    gid[id].push_back(A);
    gid[id].push_back(B);
} else if (!id1) {
     emp[A] = id2;
     gid[id2].push_back(A);
} else if (!id2) {
     emp[B] = id1;
     gid[id1].push_back(B);
} else {
   for(i = 0;i<gid[id2].size();i++) {
       emp[gid[id2][i]] = id1;
       gid[id1].push_back( gid[id2][i] );
  }
  if (manager.find(id2) != manager.end())
        manager[id1] = manager[id2].
  manager.erase(id2);
  gid.erase(id2);
}

void setManager(string A, string B) {
int id1, id2;
   if (emp.find(A) != emp.end())
         id1 = emp[A];
  else {
       ++id;
      emp[A] = id;
      id1 = id;
      gid[id].push_back(A);
  }
   if (emp.find(B) != emp.end())
         id2 = emp[B];
  else {
       ++id;
      emp[B] = id;
      id2 = id;
      gid[id].push_back(B);
  }
   manager[id2] = id1;
}

bool query(string A, string B) {
   int id1 = 0, id2 = 0;
   if (emp.find(A) != emp.end())
        id1 = emp[A];
   if (emp.find(B) != emp.end())
        id2 = emp[B];
   if (!id1 || !id2) //one of them does not exist
     return false;
  id2 = manager[id2];
  while(id2 != 0) {
     if (id2 == id1)
        return true;
     id2 = manager[id2];
 }
 return false;
}