题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2586
题目：
Problem Description
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.
Input
First 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.
Output
For 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
题意：给定一棵带权树，m次询问，两个点之间的距离。
题解：暴力DFS。
1 #include <vector>
2 #include <cstdio>
3 #include <cstring>
4 using namespace std;
5
6 const int N=12345;
7 int fa[N],ans;
8 vector < pair<int,int> > E[N];
9 vector < pair<int,int> > q[N];
10 bool root[N],vis[N];
11
12 void init(){
13 for(int i=0;i<N;i++){
14 fa[i]=i;
15 E[i].clear();
16 q[i].clear();
17 vis[i]=false;
18 root[i]=true;
19 }
20 }
21
22 int Find(int x){
23 return fa[x]==x?x:fa[x]=Find(fa[x]);
24 }
25
26 void Union(int x,int y){
27 int fx=Find(x),fy=Find(y);
28 if(fx!=fy){
29 fa[fy]=fx;
30 }
31 }
32
33 void solve(int u){
34 vis[u]=1;
35 for(int i=0;i<E[u].size();i++){
36 int v=E[u][i];
37 if(vis[v]) continue;
38 solve(v);
39 Union(u,v);
40 }
41 for(int i=0;i<q[u].size();i++){
42 int v=q[u][i];
43 if(!vis[v]) continue;
44 ans=Find(v);
45 }
46 }
47
48 int main(){
49 int t,n,m;
50 scanf("%d",&t);
51 while(t--){
52 int u,v,w;
53 init();
54 scanf("%d%d",&n,&m);
55 for(int i=1;i<n;i++){
56 scanf("%d%d%d",&u,&v,&w);
57 E[u].push_back(make_pair(v,w));
58 E[v].push_back(make_pair(u,w));
59 }
60 for(int i=1;i<=m;i++){
61 scanf("%d%d",&u,,&v);
62 Q[u].push_back(make_pair(v,i));
63 Q[v].push_back(make_pair(u,i));
64 }
65 tarjan();
66
67 }
68 return 0;
69 }
原文地址：https://www.cnblogs.com/Leonard-/p/8733105.html