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Problem Description

Avin has two trees which are not connected. He asks you to add an edge between them to make them connected while minimizing the function

, where dis(i,j) represents the number of edges of the path from i to j. He is happy with only the function value.

Input

The first line contains a number n (2<=n<=100000). In each of the following n−2 lines, there are two numbers u and v, meaning that there is an edge between u and v. The input is guaranteed to contain exactly two trees.

Output

Just print the minimum function value.

Sample Input

3

1 2

Sample Output

4

思路：

树中所有点到某个点的距离和中,到重心的距离和是最小的;

如果有两个重心,那么他们的距离和一样

两次dfs树形dp找到两颗树的重心

把两个重心连接起来 再dfs树形dp一次计算答案

计算树上任意两点的距离和，转化为计算每条边被经过的次数乘上边权

每条边被经过的次数等于这条边分隔开的 两端点分别所在的连通块的大小相乘 最后一次dfs树形dp的过程中 假设u是v的父节点， 以v为根的子树大小为sz[v] 则连接他们的边被经过的次数为(n-sz[v])*sz[v]

code：

#include
   
    #include
    
     #include
     
      #include
      
       #include
       
        #include
        
         #include
         
          using namespace std;#define ll long longconst int inf=1e9;const int maxm=1e5+5;int head[maxm],nt[maxm<<1],to[maxm<<1],cnt;int mark[maxm];int sz[maxm],son[maxm];int num;//第一个连通块的大小int size;//第二个连通块的大小int root1,root2;int n;ll ans;void init(){ memset(head,0,sizeof head); memset(mark,0,sizeof mark); cnt=1; num=0; ans=0;}void add(int x,int y){ cnt++;nt[cnt]=head[x];head[x]=cnt;to[cnt]=y;}void dfs(int x){ num++;//计算连通块大小 sz[x]=1; son[x]=0; mark[x]=1; for(int i=head[x];i;i=nt[i]){ int v=to[i]; if(mark[v])continue; dfs(v); sz[x]+=sz[v]; son[x]=max(son[x],sz[v]); }}void dfs2(int x){ sz[x]=1; son[x]=0; mark[x]=1; for(int i=head[x];i;i=nt[i]){ int v=to[i]; if(mark[v])continue; dfs2(v); sz[x]+=sz[v]; son[x]=max(son[x],sz[v]); } son[x]=max(son[x],size-sz[x]); if(son[x]
         
        
       
      
     
    
   

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