标签：启发式合并

分析

这样的三元组肯定是一个Y字形，两个点到其LCA后再向上走到当前点

考虑启发式合并，需要维护每个点到当前点的距离，记录还缺多少距离到第三个点即可

code

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<vector>
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define dep(i,a,b) for(int i=a;i>=b;i--)
#define ll long long
#define mem(x,num) memset(x,num,sizeof x)
#define reg(x) for(int i=last[x];i;i=e[i].next)
using namespace std;
inline ll read(){ll f=1,x=0;char ch=getchar();while(ch<'0'||ch>'9'){
   if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}return x*f;
}
//**********head by yjjr**********
#define pb push_back
const int maxn=1e6+6;
ll Ans=0;int n,last[maxn],cnt=0;
vector<int> son[maxn];
struct edge{
   int to,next;}e[maxn<<1];
struct state{vector<ll> dis,pr;int f;inline int size(){
   return dis.size();}void clear(){dis.clear();pr.clear();f=0;}
}*s[maxn],T[maxn];
inline bool cmp(int a,int b){
   return s[a]->size()>s[b]->size();}
void insert(int u,int v){e[++cnt]=(edge){v,last[u]};last[u]=cnt;e[++cnt]=(edge){u,last[v]};last[v]=cnt;
}
void dfs(int x,int fa){son[x].clear();reg(x){if(e[i].to==fa)continue;son[x].push_back(e[i].to);dfs(e[i].to,x);}if(!son[x].size()){s[x]=T+x;s[x]->dis.pb(1);s[x]->pr.pb(0);return;}sort(son[x].begin(),son[x].end(),cmp);s[x]=s[son[x][0]];s[x]->dis.pb(1);s[x]->f++;s[x]->pr.pb(0);s[x]->pr.pb(0);Ans+=s[x]->pr[s[x]->f];for(int i=1;i<son[x].size();i++){int v=son[x][i],f=s[v]->f,fu=s[x]->f;rep(j,0,f)Ans+=(s[v]->dis[f-j])*(s[x]->pr[j+fu+1]);rep(j,1,f)Ans+=(s[v]->pr[f+j])*(s[x]->dis[fu-(j-1)]);rep(j,0,f)s[x]->pr[j+fu+1]+=(s[v]->dis[f-j])*(s[x]->dis[fu-j-1]);rep(j,1,f)s[x]->pr[j+fu-1]+=s[v]->pr[j+f];rep(j,0,f)s[x]->dis[fu-(j+1)]+=s[v]->dis[f-j];}
}
int main()
{n=read();rep(i,1,n-1){int u=read(),v=read();insert(u,v);}dfs(1,0);cout<<Ans<<endl;return 0;
}