有一棵点数为 N 的树，以点 1 为根，且树点有边权。然后有 M 个
操作，分为三种：
操作 1 ：把某个节点 x 的点权增加 a 。
操作 2 ：把某个节点 x 为根的子树中所有点的点权都增加 a 。
操作 3 ：询问某个节点 x 到根的路径中所有点的点权和。
dfs序真是一个奥妙重重的东西
首先求出dfs序，操作1显然是一个单点修改，操作2显然是一个区间修改，我们看到我们要询问的事实上是一条链的和，我们就不能简单地进行单点修改和区间修改，事实上1到x的点在dfs序上可以视为一个栈，第一次出现是进栈，第二次出现时出栈，我们只要将进栈的点权为正，出栈的点权为负，线段树进行一波区间修改和区间查询即可。
#include <map>
#include <set>
#include <ctime>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
inline int read(){int now=;register char c=getchar();for(;!isdigit(c);c=getchar());
for(;isdigit(c);now=now*+c-'',c=getchar());return now;}
#define For(i, x, y) for(int i=x;i<=y;i++)
#define _For(i, x, y) for(int i=x;i>=y;i--)
#define Mem(f, x) memset(f,x,sizeof(f))
#define Sca(x) scanf("%d", &x)
#define Sca2(x,y) scanf("%d%d",&x,&y)
#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define Scl(x) scanf("%lld",&x);
#define Pri(x) printf("%d\n", x)
#define Prl(x) printf("%lld\n",x);
#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();
#define LL long long
#define ULL unsigned long long
#define mp make_pair
#define PII pair<int,int>
#define PIL pair<int,long long>
#define PLL pair<long long,long long>
#define pb push_back
#define fi first
#define se second
typedef vector<int> VI;
const double eps = 1e-;
const int maxn = 4e5 + ;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + ;
int N,M,K;
LL val[maxn];
struct Edge{
int to,next;
}edge[maxn * ];
int head[maxn],tot;
void init(){Mem(head,-); tot = ;}
void add(int u,int v){edge[tot].to = v;edge[tot].next = head[u];head[u] = tot++;}
PII Index[maxn * ];
PII pos[maxn];
int id = ;
void dfs(int t,int fa){
Index[++id].fi = t;
Index[id].se = ;
pos[t].fi = id;
for(int i = head[t]; ~i; i = edge[i].next){
int v = edge[i].to;
if(v == fa) continue;
dfs(v,t);
}
Index[++id].fi = t;
Index[id].se = -;
pos[t].se = id;
}
struct Tree{
int l,r;
int up,down;
LL sum,lazy;
}tree[maxn << ];
void Pushup(int t){
tree[t].sum = tree[t << ].sum + tree[t << | ].sum;
}
void Build(int t,int l,int r){
tree[t].l = l; tree[t].r = r;
tree[t].up = tree[t].down = tree[t].lazy = ;
if(l == r){
if(Index[l].se == ){
tree[t].sum = val[Index[l].fi];
tree[t].up++;
}else{
tree[t].sum = -val[Index[l].fi];
tree[t].down++;
}
return ;
}
int m = (l + r) >> ;
Build(t << ,l,m); Build(t << | ,m + ,r);
Pushup(t);
tree[t].up = tree[t << ].up + tree[t << | ].up;
tree[t].down = tree[t << ].down + tree[t << | ].down;
}
void Pushdown(int t){
if(tree[t].lazy){
tree[t << ].sum += (tree[t << ].up - tree[t << ].down) * tree[t].lazy;
tree[t << | ].sum += (tree[t << | ].up - tree[t << | ].down) * tree[t].lazy;
tree[t << ].lazy += tree[t].lazy; tree[t << | ].lazy += tree[t].lazy;
tree[t].lazy = ;
}
}
void update(int t,int l,int r,LL a){
if(l <= tree[t].l && tree[t].r <= r){
tree[t].sum += (tree[t].up - tree[t].down) * a;
tree[t].lazy += a;
return;
}
Pushdown(t);
int m = (tree[t].l + tree[t].r) >> ;
if(r <= m) update(t << ,l,r,a);
else if(l > m) update(t << | ,l,r,a);
else{
update(t << ,l,m,a);
update(t << | ,m + ,r,a);
}
Pushup(t);
}
LL query(int t,int l,int r){
if(l > r) return ;
if(l <= tree[t].l && tree[t].r <= r){
return tree[t].sum;
}
Pushdown(t);
int m = (tree[t].l + tree[t].r) >> ;
if(r <= m) return query(t << ,l,r);
else if(l > m) return query(t << | ,l,r);
else{
return query(t << ,l,m) + query(t << | ,m + ,r);
}
}
int main()
{
Sca2(N,M); init();
For(i,,N) Scl(val[i]);
For(i,,N - ){ int u,v; Sca2(u,v); add(u,v); add(v,u);}
int root = ; id = ;
dfs(root,-);
Build(,, * N);
while(M--){
int op = read();
if(op == ){
int x; Sca(x); LL a; Scl(a);
update(,pos[x].fi,pos[x].fi,a); update(,pos[x].se,pos[x].se,a);
}else if(op == ){
int x; Sca(x); LL a; Scl(a);
update(,pos[x].fi,pos[x].se,a);
}else{
int x; Sca(x);
Prl(query(,,pos[x].fi));
}
}
#ifdef VSCode
system("pause");
#endif
return ;
}