嘟嘟嘟

判断树的同构的方法就是树上哈希。

如果树是一棵有根树，那么只要从根节点出发dfs，每一个节点的哈希值等于按传统方式算出来的子树的哈希值的结果。需要注意的是，算完子树的哈希值后要先排序再加起来，因为交换两棵子树对于判断同构不影响。

对于无根树，我们应该选一个点，满足树的标号改变后树的形态没有变。所以就选重心好了。重心可能有两个（一条边的两个端点上），所以分别求一遍哈希取max（min）即可。

结果这题卡我的哈希（可能是我太菜了），最后写成ret = ((ret * base) | t[i]) % mod才过，自然溢出都不出行。

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<stack>
#include<queue>
using namespace std;
#define enter puts("")
#define space putchar(' ')
#define Mem(a, x) memset(a, x, sizeof(a))
#define In inline
typedef long long ll;
typedef unsigned long long ull;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
const ull base = 19260817;
const ull mod = 1e9 + 7;
const int maxn = 55;
inline ll read()
{ll ans = 0;char ch = getchar(), last = ' ';while(!isdigit(ch)) last = ch, ch = getchar();while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();if(last == '-') ans = -ans;return ans;
}
inline void write(ll x)
{if(x < 0) x = -x, putchar('-');if(x >= 10) write(x / 10);putchar(x % 10 + '0');
}int m, n;
struct Edge
{int nxt, to;
}e[maxn << 1];
int head[maxn], ecnt = -1;
In void addEdge(int x, int y)
{e[++ecnt] = (Edge){head[x], y};head[x] = ecnt;
}int siz[maxn], dp[maxn], G = INF;
In void dfs1(int now, int _f)
{siz[now] = 1; dp[now] = 0;for(int i = head[now], v; ~i; i = e[i].nxt){if((v = e[i].to) == _f) continue;dfs1(v, now);siz[now] += siz[v];dp[now] = max(dp[now], siz[v]);}dp[now] = max(dp[now], n - siz[now]);G = min(G, dp[now]);
}ull ans[maxn];
In ull dfs2(int now, int _f)
{ull ret = 1343247397;vector<ull> tp;for(int i = head[now], v; ~i; i = e[i].nxt){if((v = e[i].to) == _f) continue;tp.push_back(dfs2(v, now));}sort(tp.begin(), tp.end());for(int i = 0; i < (int)tp.size(); ++i) ret = ((ret * base) ^ tp[i]);return ret;
}In void init()
{Mem(head, -1); ecnt = -1;G = INF;
}int main()
{//freopen("2.in", "r", stdin);//freopen("ha.out", "w", stdout);m = read();for(int i = 1; i <= m; ++i){init();n = read();for(int j = 1; j <= n; ++j){int x = read();if(x) addEdge(x, j), addEdge(j, x);}dfs1(1, 0);for(int j = 1; j <= n; ++j)if(dp[j] == G) ans[i] = max(ans[i], dfs2(j, 0));}for(int i = 1; i <= m; ++i)for(int j = 1; j <= i; ++j)if(ans[i] == ans[j]) {write(j), enter; break;}return 0;
}