2014多校5（1008）HDU4918（点分治）

Query on the subtree

Time Limit: 16000/8000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 384    Accepted Submission(s): 133

Problem Description

bobo has a tree, whose vertices are conveniently labeled by 1,2,…,n. At the very begining, the i-th vertex is assigned with weight w _i.

There are q operations. Each operations are of the following 2 types:

Change the weight of vertex v into x (denoted as "! v x"),
Ask the total weight of vertices whose distance are no more than d away from vertex v (denoted as "? v d").

Note that the distance between vertex u and v is the number of edges on the shortest path between them.

 

Input

The input consists of several tests. For each tests:

The first line contains n,q (1≤n,q≤10 ⁵). The second line contains n integers w ₁,w ₂,…,w _n (0≤w _i≤10 ⁴). Each of the following (n - 1) lines contain 2 integers a _i,b _i denoting an edge between vertices a _i and b _i (1≤a _i,b _i≤n). Each of the following q lines contain the operations (1≤v≤n,0≤x≤10 ⁴,0≤d≤n).

 

Output

For each tests:

For each queries, a single number denotes the total weight.

 

Sample Input

    
    
    
    

     
     
     
     4 3
1 1 1 1
1 2
2 3
3 4
? 2 1
! 1 0
? 2 1
3 3
1 2 3
1 2
1 3
? 1 0
? 1 1
? 1 2
    
    
    
    

 

Sample Output

    
    
    
    

     
     
     
     3
2
1
6
6
    
    
    
    

题意：给了一棵带点权的树，支持两种操作! u w 将u的权值修改为w，? u d 询问与u的距离不超过d的点的权值和

思路：这道题拖了好久，AC以后还是很愉快的

            树分治

            对于每个分治中心center，用树状数组维护与它的距离为d的点权的和

            然后要查询与点u距离不超过d的点权值和，可以通过分治中心center来查询

            设u与center的距离为dis，则只需要统计与center距离为d-dis的点的权值和就好了，但是会将u所在的子树这部分算多余

            所以还要减去u所在子树的这部分权值和，所以还需要用树状数组维护每个子树中的点与对应的分治中心center距离为d的点的权值和，以及这个子树所对应的分治中心

            然后要查询的时候，只需要统计一遍u所在的所有子树以及对应的分治中心就可以了