Codeforces Round #225 (Div. 1) C-Propagating tree （DFS序+线段树/树状数组）

题目链接：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

题意：给出一颗有n个节点，且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v，则它的孩子节点的权值要添加-v。

题解：DFS标号后，按照深度分出奇偶层，当奇数（偶数）层+v时，对应偶数（奇数）层-v，两个线段树跑一跑就好。

线段树：

//#include <bits/stdc++.h>
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <vector>
#include <algorithm>
using namespace std;
#define INF 0x3f3f3f3f
#define LL long long
#define bug cout<<"bug"<<endl
const int MAXN = 2e5+7;
int n,m,index;
int val[MAXN],lazy[2][MAXN<<2];
vector<int> node[MAXN];
int deep,L[MAXN],R[MAXN],flag[MAXN],id[MAXN];
void dfs(int u, int fa, int poi)
{
L[u]=++deep;
id[deep]=u;
flag[u]=poi;
int siz=node[u].size();
for(int i=0; i<siz; ++i)
{
int v=node[u][i];
if(v==fa)continue;
dfs(v,u,poi^1);
}
R[u]=deep;
}
/*
void build(int l, int r, int poi)
{
if(l==r){lazy[0][poi]=lazy[1][poi]=val[id[l]];return;}
int mid=(l+r)>>1;
build(l,mid,poi<<1);
build(mid+1,r,poi<<1^1);
}*/
void push_data(int poi)
{
if(lazy[0][poi])
{
lazy[0][poi<<1]+=lazy[0][poi];
lazy[0][poi<<1^1]+=lazy[0][poi];
}
if(lazy[1][poi])
{
lazy[1][poi<<1]+=lazy[1][poi];
lazy[1][poi<<1^1]+=lazy[1][poi];
}
lazy[0][poi]=lazy[1][poi]=0;
}
void update(int l, int r, int a, int b, int poi, int f, int add)
{
if(a<=l && r<=b){lazy[f][poi]+=add;lazy[f^1][poi]-=add;return;}
push_data(poi);
int mid=(l+r)>>1;
if(mid>=a)update(l,mid,a,b,poi<<1,f,add);
if(mid<b)update(mid+1,r,a,b,poi<<1^1,f,add);
}
int query(int l, int r, int poi, int f, int p)
{
if(l==r)return lazy[f][poi];
push_data(poi);
int mid=(l+r)>>1;
if(mid>=p)return query(l,mid,poi<<1,f,p);
return query(mid+1,r,poi<<1^1,f,p);
}
int main()
{
scanf("%d%d",&n,&m);
index=0;
for(int i=1; i<=n; ++i)
scanf("%d",&val[i]);
int a,b,c;
for(int i=0; i<n-1; ++i)
{
scanf("%d%d",&a,&b);
node[a].push_back(b);
node.push_back(a);
}
deep=0;
dfs(1,0,0);
//    build(1,n,1);
for(int i=0; i<m; ++i)
{
scanf("%d%d",&a,&b);
if(a==1)
{
scanf("%d",&c);
update(1,n,L[b],R[b],1,flag[b],c);
}
else
printf("%d\n",val[b]+query(1,n,1,flag[b],L[b]));
}
return 0;
}
/*

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

*/

[b]树状数组：

//#include <bits/stdc++.h>
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <vector>
#include <algorithm>
using namespace std;
#define INF 0x3f3f3f3f
#define LL long long
#define bug cout<<"bug"<<endl
#define lowbit(a) a&(-a)
const int MAXN = 2e5+7;
int n,m,index;
int val[MAXN],lazy[2][MAXN<<2];
vector<int> node[MAXN];
int deep,L[MAXN],R[MAXN],flag[MAXN],id[MAXN];
void dfs(int u, int fa, int poi)
{
L[u]=++deep;
id[deep]=u;
flag[u]=poi;
int siz=node[u].size();
for(int i=0; i<siz; ++i)
{
int v=node[u][i];
if(v==fa)continue;
dfs(v,u,poi^1);
}
R[u]=deep;
}
void add_data(int p, int f, int data)
{
int i=p;
while(i<=n)
{
lazy[f][i]+=data;
lazy[f^1][i]-=data;
i+=lowbit(i);
}
}
int query(int p, int f)
{
int i=p,ans=0;
while(i>0)
{
ans+=lazy[f][i];
i-=lowbit(i);
}
return ans;
}
int main()
{
scanf("%d%d",&n,&m);
index=0;
for(int i=1; i<=n; ++i)
scanf("%d",&val[i]);
int a,b,c;
for(int i=0; i<n-1; ++i)
{
scanf("%d%d",&a,&b);
node[a].push_back(b);
node[b].push_back(a);
}
deep=0;
dfs(1,0,0);
for(int i=0; i<m; ++i)
{
scanf("%d%d",&a,&b);
if(a==1)
{
scanf("%d",&c);
add_data(L[b],flag[b],c);
add_data(R[b]+1,flag[b],-c);
}
else
printf("%d\n",val[b]+query(L[b],flag[b]));
}
return 0;
}
/*

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

*/