分析：这道题是树链剖分的裸题，把边的信息保存在深度大的那个节点上就行了。
一开始写的邻接表，居然TLE了。后来百度发现有人说前向星跑得比较快?我不是很明白，但是改成前向星以后的确快了很多，邻接表是T，而前向星只需要2s左右。
这个代码还是邻接表，因为我不信，又修改了一下邻接表上几个我觉得会增加常数的地方，交上去还是T了- -代码里前向星的部分被注释掉了。
/*****************************************************/
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <map>
#include <set>
#include <ctime>
#include <stack>
#include <queue>
#include <cmath>
#include <string>
#include <vector>
#include <cstdio>
#include <cctype>
#include <cstring>
#include <sstream>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
#define offcin ios::sync_with_stdio(false)
#define sigma_size 26
#define lson l,m,v<<1
#define rson m+1,r,v<<1|1
#define slch v<<1
#define srch v<<1|1
#define sgetmid int m = (l+r)>>1
#define LL long long
#define ull unsigned long long
#define mem(x,v) memset(x,v,sizeof(x))
#define lowbit(x) (x&-x)
#define bits(a) __builtin_popcount(a)
#define mk make_pair
#define pb push_back
#define fi first
#define se second
const int INF = 0x3f3f3f3f;
const LL INFF = 1e18;
const double pi = acos(-1.0);
const double inf = 1e18;
const double eps = 1e-9;
const LL mod = 1e9+7;
const int maxmat = 10;
const ull BASE = 31;
/*****************************************************/
const int maxn = 1e5 + 5;
int siz[maxn], top[maxn], son[maxn], dep[maxn], fat[maxn], tid[maxn], rid[maxn];
int seg[maxn << 2], a[maxn];
int uu[maxn], vv[maxn], ww[maxn], tim;
int N, Q, S;
std::vector<int> G[maxn];
// struct Edge {
// int to, next;
// }edge[maxn << 1];
// int head[maxn], tot;
// void addedge(int u, int v) {
// edge[tot] = (Edge){v, head[u]};
// head[u] = tot ++;
// }
void dfs1(int u, int fa, int d) {
siz[u] = 1;
fat[u] = fa;
dep[u] = d;
int sz = G[u].size();
for (int i = 0; i < sz; i ++) {
int v = G[u][i];
if (v == fa) continue;
dfs1(v, u, d + 1);
siz[u] += siz[v];
if (son[u] == -1 || siz[v] > siz[son[u]]) son[u] = v;
}
}
void dfs2(int u, int tp) {
tid[u] = ++ tim;
top[u] = tp;
if (son[u] != -1) dfs2(son[u], tp);
int sz = G[u].size();
for (int i = 0; i < sz; i ++) {
int v = G[u][i];
if (v == son[u] || v == fat[u]) continue;
dfs2(v, v);
}
}
void PushUp(int v) {
seg[v] = seg[slch] + seg[srch];
}
void build(int l, int r, int v) {
if (l == r)
seg[v] = a[l];
else {
sgetmid;
build(lson);
build(rson);
PushUp(v);
}
}
void update(int p, int val, int l, int r, int v) {
if (l == r) {
seg[v] = val;
}
else {
sgetmid;
if (p <= m) update(p, val, lson);
else update(p, val, rson);
PushUp(v);
}
}
int query(int L, int R, int l, int r, int v) {
if (L <= l && r <= R) return seg[v];
int ans = 0;
sgetmid;
if (L <= m) ans += query(L, R, lson);
if (R > m) ans += query(L, R, rson);
return ans;
}
int getv(int u, int v) {
int tp1 = top[u], tp2 = top[v];
int ans = 0;
while (tp1 != tp2) {
if (dep[tp1] < dep[tp2]) {
swap(tp1, tp2);
swap(u, v);
}
ans += query(tid[tp1], tid[u], 1, N, 1);
u = fat[tp1];
tp1 = top[u];
}
if (u == v) return ans;
if (dep[u] > dep[v]) swap(u, v);
ans += query(tid[son[u]], tid[v], 1, N, 1);
return ans;
}
int main(int argc, char const *argv[]) {
while (~scanf("%d%d%d", &N, &Q, &S)) {
mem(son, -1);
tim = 0;
for (int i = 1; i < N; i ++) {
scanf("%d%d%d", &uu[i], &vv[i], &ww[i]);
G[uu[i]].pb(vv[i]);
G[vv[i]].pb(uu[i]);
}
dfs1(1, -1, 1);
dfs2(1, 1);
for (int i = 1; i < N; i ++) {
if (dep[uu[i]] > dep[vv[i]]) swap(uu[i], vv[i]);
a[tid[vv[i]]] = ww[i];
}
build(1, N, 1);
for (int i = 0; i < Q; i ++) {
int op, x, y;
scanf("%d", &op);
if (op) {
scanf("%d%d", &x, &y);
update(tid[vv[x]], y, 1, N, 1);
}
else {
scanf("%d", &x);
int ans = getv(S, x);
printf("%d\n", ans);
S = x;
}
}
}
return 0;
}