Similarity of Subtrees 深搜，hash

https://acm.bnu.edu.cn/v3/problem_show.php?pid=52310

problem description

Define the depth of a node in a rooted tree by applying the following rules recursively:

• The depth of a root node is 0.

• The depths of child nodes whose parents are with depth d are d + 1.

Let S(T, d) be the number of nodes of T with depth d. Two rooted trees T and T ′ are similar if and only if S(T, d) equals S(T ′ , d) for all non-negative integer d. You are given a rooted tree T with N nodes. The nodes of T are numbered from 1 to N. Node 1 is the root node of T. Let Ti be the rooted subtree of T whose root is node i. Your task is to write a program which calculates the number of pairs (i, j) such that Ti and Tj are similar and i < j.

Input

The input consists of a single test case. N a1 b1 … aN−1 bN−1 The first line contains an integer N (1 ≤ N ≤ 100,000), which is the number of nodes in a tree. The following N − 1 lines give information of branches: the i-th line of them contains ai and bi , which indicates that a node ai is a parent of a node bi . (1 ≤ ai , bi ≤ N, ai ̸= bi) The root node is numbered by 1. It is guaranteed that a given graph is a rooted tree, i.e. there is exactly one parent for each node except the node 1, and the graph is connected.

Output

Print the number of the pairs (x, y) of the nodes such that the subtree with the root x and the subtree with the root y are similar and x < y. 14

Sample Input1

5

1 2

1 3

1 4

1 5

Output for the Sample Input 1

6

Sample Input2

6

1 2

2 3

3 4

1 5

5 6

Output for the Sample Input 2

2

Sample Input3

13

1 2

1 3

2 4

2 5

3 6

3 7

4 8

4 9

6 10

7 11

8 12

11 13

Output for the Sample Input 3

14

给出一棵树，求有多少对子树相似（深度相同，相同深度有相同的节点数）。

给每个叶子节点附一个值p, 父节点为子节点的值和



把p赋值成一个较大的素数就可以减少冲突，再用map映射一下

#include <iostream>
#include <stdio.h>
#include <queue>
#include <string.h>
#include <vector>
#include <map>
#include <string>
#include <set>
using namespace std;
typedef long long LL;
const LL maxn=1e6+10,p=100003,mod=1e9+7;
vector<LL>G[maxn];
LL hash_v[maxn];
map<LL,LL>mp;
map<LL,LL>::iterator it;

void dfs(LL u)
{
hash_v[u]=1;
for(LL i=0; i<G[u].size(); i++)
{
LL v=G[u][i];
dfs(v);
hash_v[u]=(hash_v[u]+hash_v[v]*p)%mod;
}
mp[hash_v[u]]++;
}
int main()
{
LL n;
LL u,v;
while(scanf("%lld",&n)!=-1)
{
for(LL i=0; i<=n; i++)
G[i].clear();
mp.clear();
for(LL i=1; i<n; i++)
{
scanf("%lld%lld",&u,&v);
G[u].push_back(v);
}
dfs(1);
LL ans=0;
for(it=mp.begin(); it!=mp.end(); it++)
ans+=it->second*(it->second-1)/2;
printf("%lld\n",ans);
}
return 0;
}