CodeForces - 778C： Peterson Polyglot （启发式合并trie树）

Posted 2021-01-31 hua-dong

Peterson loves to learn new languages, but his favorite hobby is making new ones. Language is a set of words, and word is a sequence of lowercase Latin letters.

Peterson makes new language every morning. It is difficult task to store the whole language, so Peterson have invented new data structure for storing his languages which is called broom. Broom is rooted tree with edges marked with letters. Initially broom is represented by the only vertex — the root of the broom. When Peterson wants to add new word to the language he stands at the root and processes the letters of new word one by one. Consider that Peterson stands at the vertex u. If there is an edge from u marked with current letter, Peterson goes through this edge. Otherwise Peterson adds new edge from u to the new vertex v, marks it with the current letter and goes through the new edge. Size of broom is the number of vertices in it.

In the evening after working day Peterson can‘t understand the language he made this morning. It is too difficult for bored Peterson and he tries to make it simpler. Simplification of the language is the process of erasing some letters from some words of this language. Formally, Peterson takes some positive integer p and erases p-th letter from all the words of this language having length at least p. Letters in words are indexed starting by 1. Peterson considers that simplification should change at least one word, i.e. there has to be at least one word of length at least p. Peterson tries to make his language as simple as possible, so he wants to choose p such that the size of the broom for his simplified language is as small as possible.

Peterson is pretty annoyed with this task so he asks you for help. Write a program to find the smallest possible size of the broom and integer p.

Input

The first line of input contains integer n (2 ≤ n ≤ 3·10⁵) — the size of the broom.

Next n - 1 lines describe the broom: i-th of them contains integers u_i, v_i and letter x_i — describing the edge from u_i to v_i marked with letter x_i.

Vertices are numbered from 1 to n. All x_i are lowercase latin letters. Vertex 1 is the root of the broom.

Edges describe correct broom which is made from Peterson‘s language.

Output

The first line of output should contain the minimum possible size of the broom after its simplification. The second line of output should contain integer p to choose. If there are several suitable p values, print the smallest one.

Examples

Input

5
1 2 c
2 3 a
3 4 t
2 5 t

Output

3
2

Input

16
1 2 o
2 3 f
1 4 p
4 5 i
5 6 e
6 7 c
7 8 e
4 9 r
9 10 e
10 11 t
11 12 t
12 13 y
10 14 f
14 15 i
15 16 x

Output

12
2

题意：给定一些单词的trie树，现在你可以删去其中一层（即原有的单词的某一位），使得新的trie树最小（即节点数最少）。

思路：我们需要枚举每一层，对于每一层的这些点，需要把它们各自的子树合并成新的trie树。 菜鸡我一直在想dp去做，维护LCP什么的，没想到就是合并树来做，启发式可以做到NlogN? （虽然每次用小的加到大的里面去，但是每个节点都要做一遍合并，我直观上觉得是两个log。

#include<bits/stdc++.h>
#define rep(i,a,b) for(int i=a;i<=b;i++)
using namespace std;
const int maxn=1000010;
int ch[maxn][26],cnt,a[maxn],N;
int merge(int u,int v)
{
    if(!u||!v) return u|v;  int now=++cnt;
    rep(i,0,25) ch[now][i]=merge(ch[u][i],ch[v][i]); return now;
}
void dfs(int u,int d)
{
    int now=N+1; cnt=N+1;
    rep(i,0,25) if(ch[u][i]) now=merge(now,ch[u][i]);
    a[d]+=cnt-N-1;
    rep(i,0,25) if(ch[u][i]) dfs(ch[u][i],d+1);
}
int main()
{
    int u,v; char c[3];
    scanf("%d",&N);
    rep(i,1,N-1){
        scanf("%d%d%s",&u,&v,c+1);
        ch[u][c[1]-‘a‘]=v;
    }
    dfs(1,1); int ans=0,p=0;
    rep(i,1,N) if(a[i]>ans) ans=a[i],p=i;
    printf("%d
%d
",N-ans,p);
    return 0;
}