Codeforces Round #225 (Div. 1) C. Propagating tree dfs序+树状数组

C. Propagating tree

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/383/problem/C

Description

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 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

Sample Output

3
3
0

HINT

题意

给出一颗有n个节点并一1为根节点的树，每个节点有它的权值，现在进行m次操作，操作分为添加和查询，当一个节点的权值添加val，则它的孩子节点的权值要添加-b。

题解：

dfs序+树状数组

分成两颗树做

http://blog.csdn.net/keshuai19940722/article/details/18967661

代码

//qscqesze
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define maxn 2000001
#define mod 1000000007
#define eps 1e-9
int Num;
char CH[20];
const int inf=0x3f3f3f3f;
inline ll read()
{
    int x=0,f=1;char ch=getchar();
    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;
}

//**************************************************************************************

struct node
{
    int l,r,v,d;
}node[maxn];
int n,m;
vector<int> e[maxn];
int bit[2][maxn];
int cnt;
void add(int x,int val,int *b)
{
    while(x<=n*2)
    {
        b[x]+=val;
        x+=(x&(-x));
    }
}
int get(int x,int *b)
{
    int ans=0;
    while(x>0)
    {
        ans+=b[x];
        x-=(x&(-x));
    }
    return ans;
}
void dfs(int x,int fa,int d)
{
    node[x].l=cnt++;
    node[x].d=d;
    for(int i=0;i<e[x].size();i++)
    {
        if(e[x][i]==fa)
            continue;
        dfs(e[x][i],x,1-d);
    }
    node[x].r=cnt++;
}
int main()
{
    n=read(),m=read();
    for(int i=1;i<=n;i++)
        node[i].v=read();
    for(int i=1;i<n;i++)
    {
        int a=read(),b=read();
        e[a].push_back(b);
        e[b].push_back(a);
    }
    cnt=1;
    dfs(1,-1,0);
    for(int i=0;i<m;i++)
    {
        int op=read();
        if(op==1)
        {
            int a=read(),b=read();
            add(node[a].l,b,bit[node[a].d]);
            add(node[a].r+1,-b,bit[node[a].d]);
        }
        else
        {
            int a=read();
            printf("%d
",node[a].v+get(node[a].l,bit[node[a].d])-get(node[a].l,bit[1-node[a].d]));
        }
    }
}