相信开这道题的神犇一定会树剖，但是树剖一般是维护的点权，所以对于边权，我们将该边的边权存到这条边所连的两个点中更深的那个点就好了。

注意边界！！！

我用的珂朵莉树，大概 $O (n \log {n}) \to O (n ^ 2 log {n})$，实测能过。

code：

# include <bits/stdc++.h>
# define int long long
using namespace std;
struct node
{
	long long l, r;
	mutable long long v;
	node (long long l, long long r = 0, long long v = 0) : l (l), r (r), v (v) {}
	bool operator < (const node &other) const { return l < other.l; }
};
set <node> tree;
inline set<node> :: iterator split (register long long pos)
{
	set <node> :: iterator it = tree.lower_bound (node (pos));
	if (it != tree.end () and it -> l == pos) return it; it --;
	if (it -> r < pos) return tree.end();
	register long long l = it -> l;
	register long long r = it -> r;
	register long long v = it -> v;
	tree.erase (it), tree.insert (node (l, pos - 1, v));
	return tree.insert (node (pos, r, v)).first;
}
inline void cover (int l, int r, int x)
{
	set <node> :: iterator rr = split (r + 1), ll = split (l);
	tree.erase (ll, rr); tree.insert (node (l, r, x)); return;
}
inline void add (int l, int r, int x)
{
//	cout << l << ' ' << r << endl;
	set <node> :: iterator rr = split (r + 1), ll = split (l);
	for (auto it = ll; it != rr; it ++) it -> v += x;
}
int find (int l, int r)
{
	set <node> :: iterator rr = split (r + 1);
	set <node> :: iterator ll = split (l);
	int ret = 0;
	for (auto it = ll; it != rr; it ++)
	{
		ret = max (ret, it -> v);
	}
	return ret;
}
int depth[200012], fa[200012], s[200012], son[200012];
int pos, id[200012], top[200012]; vector <int> e[200012];
int w[200012], num[200012];

int n;

inline void dfs1 (int u, int father,int dep){
	depth[u] = dep;
	fa[u] = father;
	s[u] = 1;
	int ch = -1;
	for(unsigned int i = 0; i < e[u].size (); i ++)
	{
		int v = e[u][i];
		if (v == father)
		{
			continue;
		}
		dfs1 (v, u, dep + 1);
		s[u] += s[v];
		if (s[v] > ch)
		{
			son[u] = v;
			ch = s[v];
		}
	}
}

inline void dfs2 (int u, int father)
{
	++ pos;
	id[u] = pos;
	num[pos] = w[u];
	top[u] = father;
	if (not son[u])
	{
		return;
	}
	dfs2 (son[u], father);
	for(unsigned int i = 0; i < e[u].size (); i ++)
	{
		int v = e[u][i];
		if (v == fa[u] or v == son[u])
		{
			continue;
		}
		dfs2 (v, v);
	}
}

void amend (int u, int v, int x)
{
	while (top[u] != top[v])
	{
		if (depth[top[u]] < depth[top[v]])
		{
			swap (u, v);
		}
//		cout << u << ' ' << v << endl;
		add (id[top[u]], id[u], x);
		u = fa[top[u]];
	}
	if (depth[u] > depth[v]) swap (u, v);
	if (u == v) return;
//	cout << id[u] + 1 << ' ' << id[v] << endl;
	add (id[u] + 1, id[v], x);
}

void change (int u, int v, int x)
{
	while (top[u] != top[v])
	{
		if (depth[top[u]] < depth[top[v]])
		{
			swap (u, v);
		}
		cover (id[top[u]], id[u], x);
		u = fa[top[u]];
	}
	if (u == v) return;
	if (depth[u] > depth[v]) swap (u, v);
	cover (id[u] + 1, id[v], x);
}

long long mx (int u, int v)
{
	int ret = 0;
	while (top[u] != top[v])
	{
		if (depth[top[u]] < depth[top[v]])
		{
			swap (u, v);
		}
		ret = max (ret, find (id[top[u]], id[u]));
		u = fa[top[u]];
	}
	if (depth[u] > depth[v]) swap (u, v);
	ret = max (ret, find (id[u] + 1, id[v]));
	return ret;
}
struct path
{
	int u, v, w;
}a[200012];
signed main ()
{
	ios :: sync_with_stdio (0);
	cin.tie (0), cout.tie (0);
	cin >> n;
	for (int i = 1; i < n; i ++)
	{
		cin >> a[i].u >> a[i].v >> a[i].w;
		e[a[i].u].push_back (a[i].v);
		e[a[i].v].push_back (a[i].u);
	}
	dfs1 (1, 0, 0);
	dfs2 (1, 0);
	for (int i = 1; i <= n; i ++)
	{
		int u = a[i].u, v = a[i].v, weight = a[i].w;
		if (fa[v] == u)
		{
			w[v] = weight;
			num[id[v]] = w[v];
		}
		else
		{
			w[u] = weight;
			num[id[u]] = w[u];
		}
	}
	tree.insert (node (1, n, 0));
	for (int i = 1; i <= n; i ++) cover (i, i, num[i]);
//	cout << id[2] << ' ' << num[id[2]] << endl;
	while (true)
	{
		string op;
		cin >> op;
		if (op == "Stop") break;
		if (op == "Cover")
		{
			int u, v, x;
			cin >> u >> v >> x;
			if (u == v) continue;
			change (u, v, x);
		}
		else if (op == "Change")
		{
			int k, x;
			cin >> k >> x;
			int u = a[k].u;
			int v = a[k].v;
			change (u, v, x);
		}
		else if (op == "Add")
		{
			int u, v, x;
			cin >> u >> v >> x;
			if (u == v) continue;
			amend (u, v, x);
		}
		else
		{
			int u, v;
			cin >> u >> v;
			cout << mx (u, v) << "\n";
		}
	}
}