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 1

• .

Let )

be the number of nodes of d

.

You are given a rooted tree T

with j

.

Input

The input consists of a single test case.

N

1

 

The first line contains an integer N

(i

) 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 )

of the nodes such that the subtree with the root 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\n",ans);
    return 0;
}