题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2586
思路：求两点的LCA，距离就是dis[x]+dis[y]-2*dis[lca(x,y)]，dis[i]表示的是i到树根的距离。
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <bitset>
#include <cmath>
#include <cctype>
#include <iostream>
#include <algorithm>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <sstream>
#include <iomanip>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const ll inff = 0x3f3f3f3f3f3f3f3f;
#define FOR(i,a,b) for(int i(a);i<=(b);++i)
#define FOL(i,a,b) for(int i(a);i>=(b);--i)
#define REW(a,b) memset(a,b,sizeof(a))
#define inf int(0x3f3f3f3f)
#define si(a) scanf("%d",&a)
#define sl(a) scanf("%I64d",&a)
#define sd(a) scanf("%lf",&a)
#define ss(a) scanf("%s",a)
#define mod int(1e9+7)
#define pb push_back
#define lc (d<<1)
#define Pll pair<ll,ll>
#define P pair<int,int>
#define pi acos(-1)
const int N=40008;
int deep[N],fa[N][20],n,m,dis[N],p[N];
bool vis[N];
vector<P> g[N];
void dfs(int u)
{
vis[u]=1;
for(int i=0;i<g[u].size();i++)
{
int v=g[u][i].first;
if(!vis[v]) deep[v]=deep[u]+1,dis[v]=dis[u]+g[u][i].second,dfs(v),p[v]=u;
}
}
void bz()
{
FOR(i,1,18)
FOR(j,1,n)
fa[j][i]=fa[fa[j][i-1]][i-1];
}
int lca(int x,int y)
{
if(deep[x]<deep[y]) swap(x,y);
int d=deep[x]-deep[y];
FOR(i,0,18)
if(d&(1<<i)) x=fa[x][i];
if(x==y) return x;
FOL(i,18,0)
if(fa[x][i]!=fa[y][i]) x=fa[x][i],y=fa[y][i];
return fa[x][0];
}
void ans(int x,int y)
{
int t=lca(x,y);
cout<<dis[x]+dis[y]-2*dis[t]<<endl;
}
int main()
{
cin.tie(0);
cout.tie(0);
int t;
cin>>t;
while(t--)
{
cin>>n>>m;
REW(vis,0),REW(p,0),REW(dis,0);
FOR(i,1,n) g[i].clear();
int x,y,z;
FOR(i,1,n-1)
{
si(x),si(y),si(z);
g[x].pb(P(y,z));
g[y].pb(P(x,z));
}
dfs(1);
FOR(i,1,n) if(p[i]) fa[i][0]=p[i];
bz();
while(m--)
{
si(x),si(y),ans(x,y);
}
}
return 0;
}