［codeforces383C - Propagating tree］ dfs序 +segment tree

C. Propagating tree
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
Iahub likes trees very much. Recently he discovered an interesting tree named propagating tree. The tree consists of n nodes numbered from 1 to n, each node i having an initial value ai. The root of the tree is node 1.

This tree has a special property: when a value val is added to a value of node i, the value -val is added to values of all the children of node i. Note that when you add value -val to a child of node i, you also add -(-val) to all children of the child of node i and so on. Look an example explanation to understand better how it works.

This tree supports two types of queries:

“1 x val” — val is added to the value of node x;
“2 x” — print the current value of node x.
In order to help Iahub understand the tree better, you must answer m queries of the preceding type.

Input
The first line contains two integers n and m (1 ≤ n, m ≤ 200000). The second line contains n integers a1, a2, …, an (1 ≤ ai ≤ 1000). Each of the next n–1 lines contains two integers vi and ui (1 ≤ vi, ui ≤ n), meaning that there is an edge between nodes vi and ui.

Each of the next m lines contains a query in the format described above. It is guaranteed that the following constraints hold for all queries: 1 ≤ x ≤ n, 1 ≤ val ≤ 1000.

Output
For each query of type two (print the value of node x) you must print the answer to the query on a separate line. The queries must be answered in the order given in the input.

Sample test(s)
input
5 5
1 2 1 1 2
1 2
1 3
2 4
2 5
1 2 3
1 1 2
2 1
2 2
2 4
output
3
3
0
Note
The values of the nodes are [1, 2, 1, 1, 2] at the beginning.

Then value 3 is added to node 2. It propagates and value -3 is added to it’s sons, node 4 and node 5. Then it cannot propagate any more. So the values of the nodes are [1, 5, 1,  - 2,  - 1].

Then value 2 is added to node 1. It propagates and value -2 is added to it’s sons, node 2 and node 3. From node 2 it propagates again, adding value 2 to it’s sons, node 4 and node 5. Node 3 has no sons, so it cannot propagate from there. The values of the nodes are [3, 3,  - 1, 0, 1].

题意：分层的区间加，点查；
思路：用flag染色即可

#include<iostream>
#include<stdio.h>
#include<vector>
#define lson (id*2)
#define rson (id*2+1)
#define MAXN 200005
using namespace std;
int n,m;
struct node{
    int val,lazy;
}t[MAXN*4];
int in[MAXN],out[MAXN],flag[MAXN];
int val[MAXN];
int tot=0;
vector<int> lin[MAXN];
void dfs(int x,int pre,int id)
{
    in[x]=++tot;
    flag[x]=-id;
    for(int i=0;i<lin[x].size();i++)
    {
        int v=lin[x][i];
        if(v==pre) continue ;
        dfs(v,x,-id);
    }
    out[x]=tot;
}
void push_down (int id,int l,int mid,int r)
{
    if(t[id].lazy==0) return ;
    t[lson].lazy+=t[id].lazy;
    t[rson].lazy+=t[id].lazy;
    t[id].lazy=0;
}
int query(int id,int l,int r,int pos)
{
    if(l==r)
    {
        return t[id].lazy;
    }
    int mid=(l+r)>>1;
    push_down(id,l,mid,r);
    if(pos<=mid) return query(lson,l,mid,pos);
    else return query(rson,mid+1,r,pos);
}
void add(int id,int l,int r,int L,int R,int V)
{
    if(l>R||r<L) return ;
    if(l>=L&&r<=R)
    {
        t[id].lazy+=V;
        return ;
    }
    int mid=(l+r)>>1;
    push_down(id,l,mid,r);
    add(lson,l,mid,L,R,V);
    add(rson,mid+1,r,L,R,V);
}
void solve()
{
    for(int i=1;i<=m;i++)
    {
        int aa;
        scanf("%d",&aa);
        if(aa==1) 
        {
            int bb,cc;
            scanf("%d%d",&bb,&cc);
            add(1,1,n,in[bb],out[bb],cc*flag[bb]);
        }
        else 
        {
            int bb;
            scanf("%d",&bb);
            printf("%d\n",query(1,1,n,in[bb])*flag[bb]+val[bb]);
        }
    }
}
int main()
{
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++) scanf("%d",&val[i]);
    for(int i=1;i<n;i++)
    {
        int aa,bb;
        scanf("%d%d",&aa,&bb);
        lin[aa].push_back(bb);
        lin[bb].push_back(aa);
    }
    dfs(1,0,1);
    solve();
}