Hourrank 21 Tree Isomorphism 树hash

https://www.hackerrank.com/contests/hourrank-21/challenges/tree-isomorphism

题目大意：

给出一棵树, 求有多少本质不同的子树。   N <= 19

下面给出我综合了网上一些做法后设计的hash函数(我不会证明碰撞概率)

判断两棵有根树是否相同：

将子树的Hash值从小到大排序, Hash(x) =  A * p xor Hash(son_1) mod q  * p  xor Hash(son_2) mod q ....  * p xor Hash(son_k) mod q * B mod q

判断两棵无根树是否相同：

找到重心 u, v(u 可能等于v, 即只有一个重心)  分别以它们为根求有根树Hash。不妨设 Hash(u) <= Hash(v)

Unrooted_Hash(x) = triple(n, u, v)  n为节点数。

对于本题, 只要暴力将每个子树的Unrooted_Hash 插入到set里就好了。

代码：

 //https://www.hackerrank.com/contests/hourrank-21/challenges/tree-isomorphism
 #include <cstdio>
 #include <cstring>
 #include <iostream>
 #include <cmath>
 #include <algorithm>
 #include <vector>
 #include <map>
 #include <queue>
 #include <set>
 using namespace std;

 typedef long long ll;

 #define N 20
 const int INF =  << ;
 const int P = , A = , B = ;
 const ll mod = 1e12 + ;
 const double pi = acos(-);

 vector<int> E[N], cone;
 int size[N], cnt;
 set<pair<int, pair<ll, ll> > > st;

 int DFS_SIZE(int x, int pre)
 {
     int res = ;
     for (int i = ; i < E[x].size(); ++i)
     {
         int y = E[x][i];
         if (y == pre) continue;
         res += DFS_SIZE(y, x);
     }
     return size[x] = res;
 }

 void Find_Cone(int x, int pre)
 {
     bool f = true;
     for (int i = ; i < E[x].size(); ++i)
     {
         int y = E[x][i];
         if (y == pre) continue;
         Find_Cone(y, x);
         if (size[y] > cnt / ) f = false;
     }
     if (cnt - size[x] > cnt / ) f = false;
     if (f) cone.push_back(x);
 }

 ll Hash(int x, int pre)
 {
     vector<ll> res;
     for (int i = ; i < E[x].size(); ++i)
     {
         int y = E[x][i];
         if (y == pre) continue;
         res.push_back(Hash(y, x));
     }
     sort(res.begin(), res.end());

     ll h = A;
     for (int i = ; i < res.size(); ++i) h = (h * P ^ res[i]) % mod;
     h = h * B % mod;
     return h;
 }

 int main()
 {
     //freopen("in.in", "r", stdin);
     //freopen("out.out", "w", stdout);

      int n, x, y, a[], b[];
      cin >> n;
      for (int i = ; i < n - ; ++i) cin >> a[i] >> b[i], --a[i], --b[i];
      for (int mask = ; mask < ( << n); ++mask)
      {
          for (int i = ; i < n; ++i) E[i].clear();
          cone.clear(); cnt = ;
          for (int i = ; i < n - ; ++i)
          {
              if ((mask & ( << a[i])) && (mask & ( << b[i])))
             {
                 E[a[i]].push_back(b[i]);
                 E[b[i]].push_back(a[i]);
             }
         }
         int root;
         for (int i = ; i < n; ++i) if (mask & ( << i)) root = i, ++cnt;
         if (DFS_SIZE(root, -) != cnt) continue;

         Find_Cone(root, -);
         if (cone.size() == )
         {
             ll h = Hash(cone[], -);
             st.insert(make_pair(cnt, make_pair(h, h)));
         }
         else
         {
             ll h1 = Hash(cone[], -), h2 = Hash(cone[], -);
             if (h1 > h2) swap(h1, h2);
             st.insert(make_pair(cnt, make_pair(h1, h2)));
         }
     }
     printf("%d\n", (int)st.size());
     return ;
 }