Aizu 2784 Similarity of Subtrees（树哈希）

Similarity of Subtrees

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
a2 b2
...
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

.

Sample Input 1

5
1 2
1 3
1 4
1 5

Output for the Sample Input 1

6

Sample Input 2

6
1 2
2 3
3 4
1 5
5 6

Output for the Sample Input 2

2

Sample Input 3

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
【分析】现在定义两棵树相似，仅当两棵树任意层次的节点数相等。然后给你一棵树，问你有多少棵子树是相似的。
 类似字符串，我们可以将子树哈希，然后直接比较哈希值就可以了。

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int,int>pii;
const int N = 1e5+5;
const ll p = 9901;
const ll mod = 1e9+7;
const double eps = 1e-8;
int n,m,k;
ll has[N];
vector<int>edg[N];
map<ll,ll>mp;
map<ll,ll>::iterator it;
void dfs(int u,int fa){
    has[u]=1;
    for(int v:edg[u]){
        dfs(v,u);
        has[u]=(has[u]+has[v]*p)%mod;
    }
    mp[has[u]]++;
}
int main()
{
    int u,v;
    scanf("%d",&n);
    for(int i=1;i<n;i++){
        scanf("%d%d",&u,&v);
        edg[u].push_back(v);
    }
    dfs(1,0);
    ll ans=0;
    for(it =mp.begin();it!=mp.end();it++){
        ans+=(it->second-1)*it->second/2;
    }
    printf("%lld
",ans);
    return 0;
}