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 = 1 << 30;
const int P = 10007, A = 31, B = 7;
const ll mod = 1e12 + 22333;
const double pi = acos(-1);


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 = 1;
    for (int i = 0; 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 = 0; i < E[x].size(); ++i)
    {
        int y = E[x][i];
        if (y == pre) continue;
        Find_Cone(y, x);
        if (size[y] > cnt / 2) f = false;
    }
    if (cnt - size[x] > cnt / 2) f = false;
    if (f) cone.push_back(x);
}

ll Hash(int x, int pre)
{
    vector<ll> res;
    for (int i = 0; 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 = 0; 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[20], b[20];
     cin >> n;
     for (int i = 0; i < n - 1; ++i) cin >> a[i] >> b[i], --a[i], --b[i];
     for (int mask = 1; mask < (1 << n); ++mask)
     {
         for (int i = 0; i < n; ++i) E[i].clear(); 
         cone.clear(); cnt = 0;
         for (int i = 0; i < n - 1; ++i)
         {
             if ((mask & (1 << a[i])) && (mask & (1 << b[i])))
            {
                E[a[i]].push_back(b[i]);
                E[b[i]].push_back(a[i]);
            }    
        }
        int root;
        for (int i = 0; i < n; ++i) if (mask & (1 << i)) root = i, ++cnt;
        if (DFS_SIZE(root, -1) != cnt) continue;
        
        Find_Cone(root, -1);
        if (cone.size() == 1)
        {
            ll h = Hash(cone[0], -1);
            st.insert(make_pair(cnt, make_pair(h, h)));
        }
        else
        {
            ll h1 = Hash(cone[0], -1), h2 = Hash(cone[1], -1);
            if (h1 > h2) swap(h1, h2);
            st.insert(make_pair(cnt, make_pair(h1, h2)));
        }
    }
    printf("%d
", (int)st.size());
    return 0;
}