题意：给定一棵含n个结点的生成树，共有q次操作，分为两种：
0 c：求从位置s到c的距离，然后s变成c
1 a b：把第a条边的权值变为b
题解：树链剖分
0操作就是树上区间查询。
由于该题是边权且为生成树，树链剖分只能解决点权，那我们化成点权即可。即对于边<u, v>，将离根节点远的那个点的权值赋为边权。
这样在进行树剖的时候，减去lca的点权即可。
这样1操作就是单点更新了。
#define _CRT_SECURE_NO_WARNINGS
#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<algorithm>
#include<queue>
#include<stack>
#include<cmath>
#include<vector>
#include<fstream>
#include<set>
#include<map>
#include<sstream>
#include<iomanip>
#define ll long long
#define pii pair<int, int>
using namespace std;
const int MAXN = 1e5 + 5;
int n, q, s, x[MAXN], y[MAXN], w[MAXN];
int sum[MAXN << 2], add[MAXN << 2];
int head[MAXN << 1], pre[MAXN], id[MAXN];
int siz[MAXN], son[MAXN], fa[MAXN];
int dep[MAXN], top[MAXN];
ll delta[MAXN];
struct edge {
int to;int next;
}e[MAXN << 1];
int tot, sign;
/*------------准备阶段--------------*/
void init() {
memset(head, -1, sizeof(head));memset(id, 0, sizeof(id));memset(siz, 0, sizeof(siz));memset(son, 0, sizeof(son));memset(fa, 0, sizeof(fa));memset(dep, 0, sizeof(dep));memset(top, 0, sizeof(top));memset(delta, 0, sizeof(delta));tot = sign = 0;
}
void addedge(int u, int v) {
e[tot].to = v, e[tot].next = head[u], head[u] = tot++;
}
/*--------------dfs-----------------*/
void dfs1(int u, int f) {
//1 0dep[u] = dep[f] + 1;fa[u] = f;siz[u] = 1;for (int i = head[u]; i != -1; i = e[i].next) {
int to = e[i].to;if (to == f) continue;dfs1(to, u);siz[u] += siz[to];if (siz[to] > siz[son[u]]) son[u] = to;}
}
void dfs2(int u, int Top) {
//1 1id[u] = ++sign;pre[sign] = u;top[u] = Top;if (son[u]) dfs2(son[u], Top);for (int i = head[u]; i != -1; i = e[i].next) {
int to = e[i].to;if (to == son[u] || to == fa[u]) continue;dfs2(to, to);}
}
/*--------------树状数组--------------*/
int tree[MAXN];
void update(int x, int num) {
for (; x <= n; x += x & -x)tree[x] += num;;
}
int su(int x) {
int answer = 0;for (; x > 0; x -= x & -x)answer += tree[x];return answer;
}
/*------------树链剖分-------------*/
int getSum(int x, int y) {
//查询x到y路径上所有结点和ll res = 0;while (top[x] != top[y]) {
if (dep[top[x]] < dep[top[y]]) swap(x, y);res += su(id[x]) - su(id[top[x]] - 1);x = fa[top[x]];}if (id[x] > id[y]) swap(x, y);res += su(id[y]) - su(id[x]); //减去lcareturn res;
}int main() {
init();scanf("%d%d%d", &n, &q, &s);for (int i = 1; i < n; i++) {
scanf("%d%d%d", &x[i], &y[i], &w[i]);addedge(x[i], y[i]);addedge(y[i], x[i]);}dfs1(1, 0);dfs2(1, 1);for (int i = 1; i < n; i++) {
if (fa[x[i]] == y[i]) swap(x[i], y[i]);update(id[y[i]], w[i]);}int op, a, b;while (q--) {
scanf("%d", &op);if (op == 0) {
scanf("%d", &a);printf("%d\n", getSum(s, a));s = a;}else {
scanf("%d%d", &a, &b);update(id[y[a]], b - w[a]);w[a] = b;}}return 0;
}