题目来源：
http://acm.hdu.edu.cn/showproblem.php?pid=2586
题目描述：
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
Source
Recommend
lcy
解题思路：
题目大概就是给你一颗树，然后求两个点之间的最短距离，如果用lca求的话，就是求出a和b的最近公共祖先c，然后我们可以知道a和b的最短距离就是dis【a】+dis【b】-2*dis【c】，dis的点到根节点的距离，那么就是一题裸的lca问题了。。。
代码：
#include <iostream>
#include <string>
#include <cstring>
#include <vector>
#include <queue>
#include <stack>
#include <algorithm>
#include <cmath>
#include <cctype>
#include <iomanip>
const int Maxn=40010;
using namespace std;
struct newt1{
int to,next,val;
}edge[2*Maxn];
struct new2{
int id,to,next;
}query[405];
int n,m,dis[Maxn],head_e[Maxn],head_q[Maxn],father[Maxn],ans[405],vis[Maxn],cnte,cntq;
void init()
{
for(int i=1;i<=n;i++)
{
father[i]=i;
}
memset(dis,0,sizeof(dis));
memset(head_e,-1,sizeof(head_e));
memset(head_q,-1,sizeof(head_q));
memset(vis,0,sizeof(vis));
cnte=cntq=0;
}
int fi(int x)
{
if(x==father[x])return x;
return father[x]=fi(father[x]);
}
void Union(int x,int y)
{
int u=fi(x),v=fi(y);
if(u==v)return ;
father[y]=x;
}
void addedge(int u,int v,int w)
{
edge[cnte].val=w;
edge[cnte].to=v;
edge[cnte].next=head_e[u];
head_e[u]=cnte++;
}
void addquery(int u,int v,int id)
{
query[cntq].id=id;
query[cntq].to=v;
query[cntq].next=head_q[u];
head_q[u]=cntq++;
}
void tarjan(int u)
{
vis[u]=1;
for(int i=head_e[u];i!=-1;i=edge[i].next){
int v=edge[i].to;
if(!vis[v]){
dis[v]=dis[u]+edge[i].val;
tarjan(v);
father[v]=u;
}
}
for(int i=head_q[u];i!=-1;i=query[i].next){
int v=query[i].to;
if(vis[v]){
int z=fi(v);
ans[query[i].id]=dis[u]-dis[z]-dis[z]+dis[v];
}
}
}
int main()
{
ios::sync_with_stdio(false);
int t;
cin>>t;
while(t--)
{
cin>>n>>m;
init();
for(int i=1;i<=n-1;i++)
{
int a,b,c;
cin>>a>>b>>c;
addedge(a,b,c);
addedge(b,a,c);
}
for(int i=1;i<=m;i++)
{
int a,b;
cin>>a>>b;
addquery(a,b,i);
addquery(b,a,i);
}
dis[1]=0;
tarjan(1);
for(int i=1;i<=m;i++)
cout<<ans[i]<<endl;
}
return 0;
}