好像挺神奇的，也可能是我菜。
以下称前 \(n-1\) 条边为「树边」，因为它们组成一棵树；后 \(n-1\) 条边为「回边」，因为它们由树节点回到根。
就是对于一个询问，如果 \(v\) 在 \(u\) 的子树内，发现无论如何答案都要包括 \(u\to v\) 的只经过树边的路径。那么只走这条路径一定是最优的，直接维护 \(u\) 到根仅经过树边的距离 \(d_{u}\)，答案就是 \(d_v-d_u\)。
否则必然是 \(u\) 沿着树边到子树内的点 \(t\)，然后从 \(t\) 沿回边到达 \(1\) 节点，再直接沿树边走向 \(v\)。那么这条路径的长度就是 \((d_t-d_u)+a_t+d_v\)，\(a_u\) 表示 \(u\) 到 \(1\) 的回边长度。
\(d_u+d_v\) 是定值，线段树维护 \(\min\limits_{t\in \text{sub(u)}}\{a_t+d_t\}\) 即可。修改边权相当于单点改或子树加，也可以用线段树维护。
代码很好写，跑得还挺快。
#include <bits/stdc++.h>
#define int long long
using namespace std;
namespace vbzIO {
char ibuf[(1 << 20) + 1], *iS, *iT;
#if ONLINE_JUDGE
#define gh() (iS == iT ? iT = (iS = ibuf) + fread(ibuf, 1, (1 << 20) + 1, stdin), (iS == iT ? EOF : *iS++) : *iS++)
#else
#define gh() getchar()
#endif
#define pc putchar
#define pi pair<int, int>
#define tu3 tuple<int, int, int>
#define tu4 tuple<int, int, int, int>
#define mp make_pair
#define mt make_tuple
#define fi first
#define se second
#define pb push_back
#define ins insert
#define era erase
#define clr clear
inline int read () {
char ch = gh();
int x = 0;
bool t = 0;
while (ch < '0' || ch > '9') t |= ch == '-', ch = gh();
while (ch >= '0' && ch <= '9') x = (x << 1) + (x << 3) + (ch ^ 48), ch = gh();
return t ? ~(x - 1) : x;
}
inline void write(int x) {
if (x < 0) {
x = ~(x - 1);
putchar('-');
}
if (x > 9)
write(x / 10);
putchar(x % 10 + '0');
}
}
using vbzIO::read;
using vbzIO::write;
const int maxn = 4e5 + 200;
const int inf = 1e18;
struct seg { int tg, vl; } tr[maxn << 2];
int n, m, dfc, d[maxn], b[maxn], id[maxn], sz[maxn];
vector<pi> t[maxn];
pi ed[maxn];
void dfs(int u, int fa) {
id[u] = ++dfc, sz[u] = 1;
for (auto p : t[u]) {
int v = p.fi, w = p.se;
d[v] = d[u] + w - b[u] + b[v];
dfs(v, u), sz[u] += sz[v];
}
}
#define ls x << 1
#define rs x << 1 | 1
#define mid ((l + r) >> 1)
void pushup(int x) { tr[x].vl = min(tr[ls].vl, tr[rs].vl); }
void pushtg(int x, int c) { tr[x].vl += c, tr[x].tg += c; }
void pushdown(int x) {
if (!tr[x].tg) return;
pushtg(ls, tr[x].tg), pushtg(rs, tr[x].tg);
tr[x].tg = 0;
}
void update(int l, int r, int s, int t, int c, int x) {
if (s <= l && r <= t) return pushtg(x, c);
pushdown(x);
if (s <= mid) update(l, mid, s, t, c, ls);
if (t > mid) update(mid + 1, r, s, t, c, rs);
pushup(x);
}
int query(int l, int r, int s, int t, int x) {
if (s <= l && r <= t) return tr[x].vl;
pushdown(x);
int res = inf;
if (s <= mid) res = min(res, query(l, mid, s, t, ls));
if (t > mid) res = min(res, query(mid + 1, r, s, t, rs));
return res;
}
signed main() {
n = read(), m = read();
for (int i = 1, u, v, w; i <= n - 1; i++) {
u = read(), v = read(), w = read();
t[u].pb(mp(v, w)), ed[i] = mp(v, w);
}
for (int i = 1, u, v, w; i <= n - 1; i++) {
u = read(), v = read(), w = read();
ed[i + n - 1] = mp(u, w), b[u] = w;
}
dfs(1, 0);
for (int i = 2; i <= n; i++) update(1, n, id[i], id[i], d[i], 1);
while (m--) {
int op = read();
if (op == 2) {
int u = read(), v = read();
int duv = (query(1, n, id[v], id[v], 1) - b[v]) - (query(1, n, id[u], id[u], 1) - b[u]);
if (id[v] >= id[u] && id[v] <= id[u] + sz[u] - 1) write(duv), puts("");
else write(duv + query(1, n, id[u], id[u] + sz[u] - 1, 1)), puts("");
} else {
int p = read(), w = read();
if (p <= n - 1) update(1, n, id[ed[p].fi], id[ed[p].fi] + sz[ed[p].fi] - 1, w - ed[p].se, 1), ed[p].se = w;
else update(1, n, id[ed[p].fi], id[ed[p].fi], w - b[ed[p].fi], 1), b[ed[p].fi] = w;
}
}
return 0;
}