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"<<endlconst 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[b].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 51 2 1 1 21 21 32 42 51 2 31 1 22 12 22 4*/

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 51 2 1 1 21 21 32 42 51 2 31 1 22 12 22 4*/