You are given a tree T consisting of n vertices and n−1 edges. Each edge of the tree is associated
with a lowercase English letter ci
.
You are given a string s consisting of lowercase English letters. Your task is to find a simple path
in the tree such that the string formed by concatenation of letters associated with edges of this
path contains string s as a subsequence, or determine that there exists no such simple path.
Input
The first line of input contains two positive integers n and m (2 ≤ n ≤ 5 · 105
, 1 ≤ m ≤ n − 1),
the number of vertices in the tree and the length of the string s.
The following n − 1 lines contain triples ui
, vi
, ci (1 ≤ ui
, vi ≤ n, ui ̸= vi
, ci
is a lowercase English
letter), denoting an edge (ui
, vi) associated with letter ci
.
The last line contains a string s (|s| = m) consisting of lowercase English letters.
Output
If the desired path exists, output its endpoints a and b. Otherwise, output “-1 -1”. If there are
several possible answers, you are allowed to output any of them.
Example
input
9 3
1 2 a
2 3 b
2 4 a
4 5 b
4 6 c
6 7 d
6 8 a
8 9 b
acb
output
8 3

題意:

給你n-1行（wa在這裡兩次）每行包含三個數：兩條相連的點和這個邊的權值（一個字元），最後給你一個長為m的字串，讓你求任意兩個端點使得這兩個端點的路徑上的字串包含這個字元8 3 和3 8是不一樣的。

題解：

先建樹，接下來從1開始dfs，找到葉子節點回溯的時候看是否首尾的長度已經有m了，然後再把兒子的值更新給父親。

#include<bits/stdc++.h>
using namespace std;
#define pa pair<int,int>
#define mp(a,b) make_pair(a,b)
const int N=5e5+5;
struct node
{
    int to,next,val;
}e[N<<1];
int head[N],cnt,n,m;
char ss[N];
pa sta[N],en[N],ans;
void add(int x,int y,int w)
{
    e[cnt].to=y;
    e[cnt].next=head[x];
    e[cnt].val=w;
    head[x]=cnt++;
}
char s[2];
void dfs(int son,int fa)
{
    sta[son]=en[son]=mp(0,son);
    //if(ans.first!=-1&&ans.second!=-1)
        //return ;
    for(int i=head[son];~i;i=e[i].next)
    {
        int ne=e[i].to;
        if(ne==fa)
            continue;
        dfs(ne,son);
        if(sta[ne].first<m&&e[i].val==ss[sta[ne].first])
            sta[ne].first++;
        if(en[ne].first<m&&e[i].val==ss[m-1-en[ne].first])
            en[ne].first++;
        if(sta[son].first+en[ne].first>=m)
            ans=mp(sta[son].second,en[ne].second);
        if(sta[ne].first+en[son].first>=m)
            ans=mp(sta[ne].second,en[son].second);
        sta[son]=max(sta[son],sta[ne]);
        en[son]=max(en[son],en[ne]);
    }
}
int main()
{
    memset(head,-1,sizeof(head));
    scanf("%d%d",&n,&m);
    int l,r;
    for(int i=1;i<n;i++)
    {
        scanf("%d%d%s",&l,&r,s);
        add(l,r,(int)s[0]),add(r,l,(int)s[0]);
    }
    scanf("%s",ss);
    ans=mp(-1,-1);
    dfs(1,0);
    printf("%d %d\n",ans.first,ans.second);
    return 0;
}