倍增法(在线)
预处理O(nlogn),每次查询O(logn)

int h[N], e[M], ne[M], w[M], idx; //memset(h, -1, sizeof h);
int depth[N], fa[N][21];
int dist[N];

void add(int a, int b, int c) //添加一条由a指向b的边,权重为c
{
    w[idx] = c;
    e[idx] = b;
    ne[idx] = h[a];
    h[a] = idx++;
}

void bfs(int root)
{
    depth[root] = 1;
    queue<int> q;
    q.push(root);
    while (q.size())
    {
        int u = q.front();
        q.pop();
        for (int i = h[u]; ~i; i = ne[i])
        {
            int v = e[i];
            if (depth[v])
            {
                continue;
            }
            depth[v] = depth[u] + 1;
            dist[v] = dist[u] + w[i];
            q.push(v);
            fa[v][0] = u;
            for (int k = 1; k <= 20; k++)
            {
                fa[v][k] = fa[fa[v][k - 1]][k - 1];
            }
        }
    }
}

int lca(int a, int b)//返回a,b的最近公共祖先
{
    if (depth[a] < depth[b])
    {
        swap(a, b);
    }
    for (int k = 20; k >= 0; k--)
    {
        if (depth[fa[a][k]] >= depth[b])
        {
            a = fa[a][k];
        }
    }
    if (a == b)
    {
        return a;
    }
    for (int k = 20; k >= 0; k--)
    {
        if (fa[a][k] != fa[b][k])
        {
            a = fa[a][k];
            b = fa[b][k];
        }
    }
    return fa[a][0];
}

Tarjan(离线)
O(n+m)

int h[N], e[M], ne[M], w[M], idx; //memset(h, -1, sizeof h);
int dist[N];
int p[N];
int res[M];
int st[N];
vector<PII> query[N]; // first存查询的另外一个点，second存查询编号

void add(int a, int b, int c) //添加一条由a指向b的边,权重为c
{
    w[idx] = c;
    e[idx] = b;
    ne[idx] = h[a];
    h[a] = idx++;
}

void dfs(int u, int father)
{
    for (int i = h[u]; ~i; i = ne[i])
    {
        int v = e[i];
        if (v == father)
        {
            continue;
        }
        dist[v] = dist[u] + w[i];
        dfs(v, u);
    }
}

int find(int x) //返回x的祖宗节点
{
    return x == p[x] ? x : p[x] = find(p[x]);
}

void tarjan(int u)
{
    st[u] = 1;
    for (int i = h[u]; ~i; i = ne[i])
    {
        int v = e[i];
        if (!st[v])
        {
            tarjan(v);
            p[v] = u;
        }
    }
    for (auto item : query[u])
    {
        int y = item.first, id = item.second;
        if (st[y] == 2)
        {
            int anc = find(y);
            res[id] = dist[u] + dist[y] - 2 * dist[anc];
        }
        st[u] = 2;
    }
}

int main()
{
    std::ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    memset(h, -1, sizeof h);
    cin >> n >> m;
    for (int i = 1; i < n; i++)
    {
        int a, b, c;
        cin >> a >> b >> c;
        add(a, b, c), add(b, a, c);
    }
    for (int i = 0; i < m; i++) //存储提问
    {
        int a, b;
        cin >> a >> b;
        if (a != b)
        {
            query[a].push_back({b, i});
            query[b].push_back({a, i});
        }
    }
    for (int i = 1; i <= n; i++) //初始化并查集
    {
        p[i] = i;
    }
    dfs(1, -1); //初始化
    tarjan(1);
    for (int i = 0; i < m; i++) //输出结果
    {
        cout << res[i] << endl;
    }
    return 0;
}