Problem
\ (It is known that there are n nodes with n−1 edges, forming a tree structure \) .
\ (Given a root node k, each node has a weight, and the weight of node i is vi \)
\ (For m operations, there are two types of operations: \)
\ (1 \ space a \ space x: means to add the weight of node a to x \)
\ (2 \ space a: means sum of all nodes in the subtree of node a (including node a itself) \)
Solution
\ (There is a conclusion: the timestamp dfn of a node and all its subtree nodes is continuous \)
\ (We use a dfn timestamp, time1 when entering, time2 \ after traversing all the subtree nodes)
\ (Then the range of this node and all its sub-tree nodes is a continuous time1-time2, then you can use the tree array to maintain \)
\ (That is, through dfn time stamp (dfs order), convert the tree into a continuous linear segment \)
#include<bits/stdc++.h>
#define IOS ios::sync_with_stdio(0); cin.tie(0);
using namespace std;
typedef long long ll;
const int maxn = 1e6+10;
ll val[maxn];
ll valu[maxn];
vector<int>E[maxn];
int dfn[maxn],ed[maxn];
int cnt;
int n,m,k;
int lowbit(int x){
return x&-x;
}
void add(int x,ll w){
while(x<=n){
val[x] += w;
x += lowbit(x);
}
}
ll query(int x){
ll res = 0;
while(x){
res += val[x];
x -= lowbit(x);
}
return res;
}
void dfs(int p,int fa){
dfn[p] = cnt++; add(dfn[p],valu[p]);
for(int i=0;i<(int)E[p].size();i++){
int v = E[p][i];
if(v!=fa){
dfs(v,p);
}
}
ed[p] = cnt-1;
}
int main(){
IOS
cin>>n>>m>>k;
for(int i=1;i<=n;i++) cin>>valu[i];
for(int i=1;i<n;i++){
int u,v;
cin>>u>>v;
E[u].push_back(v);
E[v].push_back(u);
}
cnt = 1;
dfs(k,0);
for(int i=1;i<=m;i++){
int op,p;
cin>>op>>p;
if(op==1){
ll w;
cin>>w;
add(dfn[p],w);
}else{
cout<<query(ed[p])-query(dfn[p]-1)<<endl;
}
}
return 0;
}