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&＃xff0c;每个节点有它的权值&＃xff0c;现在进行m次操作&＃xff0c;操作分为添加和查询&＃xff0c;当一个节点的权值添加val&＃xff0c;则它的孩子节点的权值要添加-b。

题解&＃xff1a;

方法一&＃xff1a;dfs序&＃43;树状数组

学过或者没学过树链剖分的小朋友应该知道dfs序可以使得任意节点的子树变成一段连续的区间&＃xff0c;这样就可以进行区间修改了。

但是这道题对子树层数的不同&＃xff0c;操作也不一样&＃xff0c;确实很麻烦呀。

我也是一脸懵逼&＃xff0c;看完题解才发现原来如此神奇&＃xff01;&＃xff01;

其实我们可以把树分层&＃xff0c;分成奇数层和偶数层&＃xff08;根据到根节点的距离&＃xff09;&＃xff0c;然后建两个树状数组&＃xff0c;如果是对奇数层的点进行操作那么就给第一个树状数组相应的区间加上val,如果是偶数层就更新第二个树状数组。

这样对于某个节点i&＃xff08;假设是奇数层&＃xff09;&＃xff0c;那么此刻权值就是树状数组&＃xff11;&＃xff0d;树状数组&＃xff12;。

感觉这种分层的思想很是神奇。

方法二&＃xff1a;dfs序&＃xff0b;线段树

第二天遇到另外一题 https://www.cnblogs.com/mmmqqdd/p/10844365.html ,突然发现这道题就是那题的简化版&＃xff0c;大家可以先看看那题&＃xff0c;

因为这道题用线段树实在是有点杀鸡用牛刀了&＃xff0c;但是思想和不错哟。

我在这也简单的讲一下做法吧&＃xff1a;

就是线段树每个节点都标记是奇数层还是偶数层&＃xff0c;奇数层为&＃xff0d;&＃xff11;&＃xff0c;偶数层为&＃xff11;&＃xff0c;然后如果处理奇数层就区间-val&＃xff0c;偶数层就&＃43;val,其他的和线段树一样&＃xff0c;最后求答案时&＃xff0c;只要把最后的值×&＃xff11;或&＃xff0d;&＃xff11;&＃xff08;奇数层&＃xff0d;&＃xff11;&＃xff0c;偶数层&＃xff11;&＃xff09;。

这个题目完全可以问一棵子数的权值和&＃xff0c;这样像一棵正常的线段树了。

代码&＃xff11;&＃xff0d;&＃xff0d;树状数组:

#include
using namespace std;
#define ll long long
#define N 400050
int n,m,cnt,w[N],a[N],c[2][N];
int dp[N],rk[N],kth[N],size[N];
int tot,last[N];
struct Edge{int from,to,s;}edges[N<<1];
templatevoid read(T&x)
{ll k&＃61;0; char c&＃61;getchar();x&＃61;0;while(!isdigit(c)&&c!&＃61;EOF)k^&＃61;c&＃61;&＃61;&＃39;-&＃39;,c&＃61;getchar();if (c&＃61;&＃61;EOF)exit(0);while(isdigit(c))x&＃61;x*10&＃43;c-&＃39;0&＃39;,c&＃61;getchar();x&＃61;k?-x:x;
}
void read_char(char &c)
{while(!isalpha(c&＃61;getchar())&&c!&＃61;EOF);}
void AddEdge(int x,int y)
{edges[&＃43;&＃43;tot]&＃61;Edge{x,y,last[x]};last[x]&＃61;tot;
}
void dfs(int x,int pre)
{dp[x]&＃61;1-dp[pre];rk[x]&＃61;&＃43;&＃43;cnt;kth[cnt]&＃61;x;size[x]&＃61;1;for(int i&＃61;last[x];i;i&＃61;edges[i].s){Edge &e&＃61;edges[i];if (e.to&＃61;&＃61;pre)continue;dfs(e.to,x);size[x]&＃43;&＃61;size[e.to];}
}
void update(int *c,int x,int tt)
{while(x<&＃61;n){c[x]&＃43;&＃61;tt;x&＃43;&＃61;x&-x;}}
int get_sum(int *c,int x)
{int ans&＃61;0;while(x){ans&＃43;&＃61;c[x];x-&＃61;x&-x;}return ans;}
int main()
{
#ifndef ONLINE_JUDGEfreopen("aa.in","r",stdin);
#endifread(n);read(m);for(int i&＃61;1;i<&＃61;n;i&＃43;&＃43;)read(w[i]);for(int i&＃61;1;i<&＃61;n-1;i&＃43;&＃43;){int x,y;read(x); read(y);AddEdge(x,y);AddEdge(y,x);}dfs(1,0);for(int i&＃61;1;i<&＃61;m;i&＃43;&＃43;){int id,x,val;read(id);if (id&＃61;&＃61;1){read(x); read(val);update(c[dp[x]],rk[x],val);update(c[dp[x]],rk[x]&＃43;size[x],-val);}if (id&＃61;&＃61;2){read(x);int ans&＃61;get_sum(c[dp[x]],rk[x])-get_sum(c[1-dp[x]],rk[x]);ans&＃43;&＃61;w[x];printf("%d\n",ans);}}}

代码&＃xff12;&＃xff0d;&＃xff0d;线段树:

#include
using namespace std;
#define ll long long
#define N 200050
int n,m,w[N];
int tot,last[N];
int cnt,rk[N],kth[N],d[N],size[N];
struct Tree{int l,r,c;}tr[N<<2];
struct Edge{int from,to,s;}edges[N<<1];
void AddEdge(int x,int y)
{edges[&＃43;&＃43;tot]&＃61;Edge{x,y,last[x]};last[x]&＃61;tot;
}
templatevoid read(T&x)
{ll k&＃61;0; char c&＃61;getchar();x&＃61;0;while(!isdigit(c)&&c!&＃61;EOF)k^&＃61;c&＃61;&＃61;&＃39;-&＃39;,c&＃61;getchar();if (c&＃61;&＃61;EOF)exit(0);while(isdigit(c))x&＃61;x*10&＃43;c-&＃39;0&＃39;,c&＃61;getchar();x&＃61;k?-x:x;
}
void read_char(char &c)
{while(!isalpha(c&＃61;getchar())&&c!&＃61;EOF);}
void dfs(int x,int pre)
{rk[x]&＃61;&＃43;&＃43;cnt;kth[cnt]&＃61;x;d[x]&＃61;-d[pre];size[x]&＃61;1;for(int i&＃61;last[x];i;i&＃61;edges[i].s){Edge &e&＃61;edges[i];if (e.to&＃61;&＃61;pre)continue;dfs(e.to,x);size[x]&＃43;&＃61;size[e.to];}
}
void push_down(int x)
{Tree &a&＃61;tr[x<<1],&b&＃61;tr[x<<1|1];a.c&＃43;&＃61;tr[x].c;b.c&＃43;&＃61;tr[x].c;tr[x].c&＃61;0;
}
void bt(int x,int l,int r)
{tr[x].l&＃61;l; tr[x].r&＃61;r; if (l&＃61;&＃61;r) return;int mid&＃61;(l&＃43;r)>>1;bt(x<<1,l,mid);bt(x<<1|1,mid&＃43;1,r);
}
void update(int x,int l,int r,int tt)
{if (l<&＃61;tr[x].l&&tr[x].r<&＃61;r){tr[x].c&＃43;&＃61;tt;return;}int mid&＃61;(tr[x].l&＃43;tr[x].r)>>1;if (l<&＃61;mid)update(x<<1,l,r,tt);if (mid}
int query(int x,int p)
{if (p<&＃61;tr[x].l&&tr[x].r<&＃61;p)return tr[x].c*d[kth[tr[x].l]];int mid&＃61;(tr[x].l&＃43;tr[x].r)>>1;push_down(x);if (p<&＃61;mid)return query(x<<1,p);else return query(x<<1|1,p);
}
int main()
{
#ifndef ONLINE_JUDGEfreopen("aa.in","r",stdin);
#endifread(n); read(m);for(int i&＃61;1;i<&＃61;n;i&＃43;&＃43;)read(w[i]);for(int i&＃61;1;i<&＃61;n-1;i&＃43;&＃43;){int x,y;read(x); read(y);AddEdge(x,y);AddEdge(y,x);}d[0]&＃61;1;dfs(1,0);bt(1,1,n);for(int i&＃61;1;i<&＃61;m;i&＃43;&＃43;){int id,x,tt;read(id); read(x);if (id&＃61;&＃61;1){read(tt);update(1,rk[x],rk[x]&＃43;size[x]-1,tt*d[x]);}if (id&＃61;&＃61;2)printf("%d\n",w[x]&＃43;query(1,rk[x]));}
}