POJ - 2763 动态lca加dfs序树状数组

一开始用树链剖分做的，后来看题解有一个稍微代码量少一点而且不用一错误的方法，lca加上树状数组。
对一条边变成w等于把深的点包括自己和子树全部加上w，即在dfs中l,r区间更新即可。
第一次写了st表求lca，有点不顺手，记录一下模板。

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<cstdlib>
#include<climits>
#include<stack>
#include<vector>
#include<queue>
#include<set>
#include<bitset>
#include<map>
//#include<regex>
#include<cstdio>
#define up(i,a,b)  for(int i=a;i<b;i++)
#define dw(i,a,b)  for(int i=a;i>b;i--)
#define upd(i,a,b) for(int i=a;i<=b;i++)
#define dwd(i,a,b) for(int i=a;i>=b;i--)
//#define local
typedef long long ll;
const double esp = 1e-6;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int inf = 1e9;
using namespace std;
ll read()
{
	char ch = getchar(); ll x = 0, f = 1;
	while (ch<'0' || ch>'9') { if (ch == '-')f = -1; ch = getchar(); }
	while (ch >= '0' && ch <= '9') { x = x * 10 + ch - '0'; ch = getchar(); }
	return x * f;
}
typedef pair<int, int> pir;
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define lrt root<<1
#define rrt root<<1|1
const int N = 1e5 + 10;
struct node { int to, next, wi; }edge[N<<2];
int head[N];
int n, q, s;
int cnt;
int rk[N], r[N], l[N],dis[N];
int ver[2 * N], fi[2 * N], dep[2 * N];
ll sum[N << 1];
int cnt1, cnt2;
int dp[N<<1][30];
int lowbit(int i)
{
	return i & (-i);
}
void addedge(int u, int v, int w)
{
	edge[cnt].to = v;
	edge[cnt].wi = w;
	edge[cnt].next = head[u];
	head[u] = cnt++;
}
void dfs(int u, int f,int d)
{
	ver[++cnt1] = u; dep[cnt1] = d; fi[u] = cnt1;
	l[u] = ++cnt2;
	for (int i = head[u]; ~i; i = edge[i].next)
	{
		int v = edge[i].to;
		if (v == f)continue;
		dfs(v, u, d + 1);
		dis[v] = edge[i].wi;
		ver[++cnt1] = u; dep[cnt1] = d;
	}
	r[u] = cnt2;
}
void stinit()
{
	upd(i, 0, 2 * n)dp[i][0] = i;
	for (int j = 1; j <= 30; j++)
	{
		for (int i = 1; i + (1 << j)-1 <= 2 * n; i++)
		{
			int t1 = dp[i][j - 1]; int t2 = dp[i + (1 << (j - 1))][j - 1];
			dp[i][j] = dep[t1] < dep[t2] ? t1 : t2;
		}
	}
}
int stquerry(int x, int y)
{
	if (x > y)swap(x, y);
	int k = int(log2(y - x + 1));
	int t1 = dp[x][k]; int t2 = dp[y - (1 << k) + 1][k];
	return dep[t1] < dep[t2] ? ver[t1] : ver[t2];
}
void add(int pos, ll val)
{
	while (pos <= n)
	{
		sum[pos] += val;
		pos += lowbit(pos);
	}
}
ll querry(int x)
{
	ll ans = 0;
	while (x)
	{
		ans += sum[x];
		x -= lowbit(x);
	}
	return ans;
}
vector<node>vec;
int main()
{
	n = read(); q = read(); s = read();
	memset(head, -1, sizeof(head));
	int u, v, w;
	upd(i, 1, n - 1)
	{
		u = read(), v = read(); w = read();
		addedge(u, v, w);
		addedge(v, u, w);
		vec.push_back(node{ u,v,w });
	}
	dfs(1, 0, 1);
	stinit();
	upd(i, 1, n)
	{
		add(l[i], dis[i]);
		add(r[i] + 1, -dis[i]);
	}
	int op=0;
	while (q--)
	{
		scanf("%d", &op);
		if (op == 0)
		{
			u = read();
			printf("%lld\n", querry(l[u]) + querry(l[s]) - 2ll * querry(l[stquerry(fi[u], fi[s])]));	
			s = u;
		}
		else
		{
			u = read(); w = read();
			u--;
			int t1 = vec[u].next; int t2 = vec[u].to;
			if (dep[fi[t1]] < dep[fi[t2]])
			{
				swap(t1, t2);
			}
			add(l[t1], -vec[u].wi);
			add(r[t1] + 1, +vec[u].wi);
			add(l[t1], w);
			add(r[t1] + 1, -w);
			vec[u].wi = w;
		}
	}
	return 0;
}