JAG Asia 2016-Similarity of Subtrees(BFS+哈希)

Similarity of Subtrees

时间限制: 1 Sec  内存限制: 128 MB

题目描述

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.

输入

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.

输出

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.

样例输入

5
1 2
1 3
1 4
1 5

样例输出

6

#include<bits/stdc++.h>
#pragma GCC optimize(3)
using namespace std;
typedef long long ll;
const int maxn=1e6+7;
const ll prime=1e5+7;
const ll mod=1e9+7;
map<ll,ll>MAP;
ll Hash_val[maxn];
vector<int>vct[maxn];
void bfs(int a)
{
    Hash_val[a]=1;
    int siz=vct[a].size();
    for(int i=0;i<siz;++i)
    {
        int b=vct[a][i];
        bfs(b);
        Hash_val[a]=(Hash_val[a]+Hash_val[b]*prime)%mod;
    }
    MAP[Hash_val[a]]++;
}
int main()
{
    int n;
    scanf("%d",&n);
    for(int i=1;i<n;++i)
    {
        int a,b;
        scanf("%d%d",&a,&b);
        vct[a].push_back(b);
    }
    ll ans=0;
    bfs(1);
    for(map<ll,ll>::iterator it=MAP.begin();it!=MAP.end();++it)
    {
        ans+=(it->second-1)*(it->second)/2;
    }
    printf("%lld
",ans);
    return 0;
}