HDU-2586 How far away？

There are n houses in the village and some bidirectional roads connecting them. Every day peole always like to ask like this "How far is it if I want to go from house A to house B"? Usually it hard to answer. But luckily int this village the answer is always unique, since the roads are built in the way that there is a unique simple path("simple" means you can't visit a place twice) between every two houses. Yout task is to answer all these curious people.

InputFirst line is a single integer T(T<=10), indicating the number of test cases.
  For each test case,in the first line there are two numbers n(2<=n<=40000) and m (1<=m<=200),the number of houses and the number of queries. The following n-1 lines each consisting three numbers i,j,k, separated bu a single space, meaning that there is a road connecting house i and house j,with length k(0<k<=40000).The houses are labeled from 1 to n.
  Next m lines each has distinct integers i and j, you areato answer the distance between house i and house j.OutputFor each test case,output m lines. Each line represents the answer of the query. Output a bland line after each test case.Sample Input

2
3 2
1 2 10
3 1 15
1 2
2 3

2 2
1 2 100
1 2
2 1

Sample Output

10
25
100
100

 1 #include<iostream>
 2 #include<stdio.h>
 3 #include<vector>
 4 #include<algorithm>
 5 #include<string.h>
 6 using namespace std;
 7 int ans[250],vis[40500];
 8 vector< pair <int ,int > > vec[40500];
 9 vector <pair <int ,int > > query[40500];
10 struct Node{
11     int f;
12     int len;
13 } pre[40500];
14 Node Find(int x)
15 {
16     if(pre[x].f==x)
17     {
18         return pre[x];
19     }
20     else
21     {
22         Node now;
23         now=Find(pre[x].f);
24         pre[x].len+=now.len;
25         pre[x].f=now.f;
26         return pre[x];
27     }
28 }
29 int dfs(int u,int fa,int length)
30 {
31     pre[u].f=u;
32     pre[u].len=0;
33     vis[u]=1;
34     for(int i=0;i<vec[u].size();i++)
35     {
36         int v=vec[u][i].first;
37         int len=vec[u][i].second;
38         if(v==fa)   continue;
39         dfs(v,u,len);
40     }
41     for(int i=0;i<query[u].size();i++)
42     {
43         int v=query[u][i].first;
44         int id=query[u][i].second;
45         if(vis[v]==1)
46         {
47             ans[id]=Find(v).len+Find(u).len;
48         }
49     }
50     pre[u].f=fa;
51     pre[u].len+=length;
52 }
53 int main()
54 {
55     int T;
56     cin>>T;
57     while(T--)
58     {
59         int n,q,x,y,k;
60         cin>>n>>q;
61         for(int i=0;i<n-1;i++)
62         {
63             cin>>x>>y>>k;
64             vec[x].push_back({y,k});
65             vec[y].push_back({x,k});
66             vis[y]=1;
67         }
68         for(int i=0;i<q;i++)
69         {
70             cin>>x>>y;
71             query[x].push_back({y,i});
72             query[y].push_back({x,i});
73         }
74         for(int i=1;i<=n;i++)
75         {
76             if(vis[i]==0)
77             {
78                 dfs(i,i,0);
79                 break;
80             }
81         }
82         for(int i=0;i<q;i++)
83             cout<<ans[i]<<endl;
84         for(int i=0;i<=n;i++)
85         {
86             vec[i].clear();
87             query[i].clear();
88         }
89         memset(pre,0,sizeof(pre));
90         memset(vis,0,sizeof(vis));
91     }
92     return 0;
93 }