Codeforces 383C Propagating tree(树状数组)

参考：http://blog.csdn.net/accelerator_/article/details/18654307
http://codeforces.com/blog/entry/10476

C. Propagating tree
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard 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].

You can see all the definitions about the tree at the following link: http://en.wikipedia.org/wiki/Tree_(graph_theory)

题意：n个结点的树，以1为根，可以往结点添加值，添加之后他的子孩子会添加负的该值，直到树的叶子结点为止。有询问和添加值的操作

思路：树状数组，先进行一遍dfs，把每个结点对应的孩子的区间l，r记录下来，然后进行树状数组的区间更新，询问的时候就计算1-l的和，因为这些都为该结点的父亲结点，然后开两个数组，一个作为正一个作为负。

代码：

#include<bits/stdc++.h>using namespace std;#define LL long long#define N 200010#define pii pair<int,int>int n,m;int bit[2][N];void add(int x,int val,int *bit){    while(x<=2*n){        bit[x]+=val;        x+=(x&(-x));    }}int get(int x,int *bit){    int ret=0;    while(x){        ret+=bit[x];        x-=(x&(-x));    }    return ret;}struct Node{    int l,r;    int tp,value;}node[N];int fst[N],vv[N<<1],nxt[N<<1],e;void add(int u,int v){    vv[e]=v;nxt[e]=fst[u];fst[u]=e++;}int tot;void dfs(int u,int fa,int d){    node[u].l=tot++;node[u].tp=d;    for(int i=fst[u];~i;i=nxt[i]){        int v=vv[i];        if(v==fa)continue;        dfs(v,u,!d);    }    node[u].r=tot++;}void init(){    scanf("%d%d",&n,&m);    tot=1;    memset(fst,-1,sizeof(fst));e=0;    for(int i=1;i<=n;++i)scanf("%d",&node[i].value);    for(int i=1;i<n;++i){        int u,v;        scanf("%d%d",&u,&v);        add(u,v);add(v,u);    }    dfs(1,1,0);    for(int i=1;i<=n;++i){        printf("%d %d\n",node[i].l,node[i].r);    }puts("");}void solve(){    for(int i=1;i<=m;++i){        int op;scanf("%d",&op);        if(op==1){            int x,val;scanf("%d%d",&x,&val);            add(node[x].l,val,bit[node[x].tp]);            add(node[x].r+1,-val,bit[node[x].tp]);        }        else{            int x;scanf("%d",&x);            printf("%d\n",node[x].value+get(node[x].l,bit[node[x].tp])-get(node[x].l,bit[!node[x].tp]));        }    }}int main(){    //freopen("in.txt","r",stdin);    init();    solve();}