Tree
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 700 Accepted Submission(s): 145
Problem Description
You are given a tree (an acyclic undirected connected graph) with N nodes. The tree nodes are numbered from 1 to N
There are N - 1 edges numbered from 1 to N - 1.
Each node has a value and each edge has a value. The initial value is 0.
There are two kind of operation as follows:
● ADD1 u v k: for nodes on the path from u to v, the value of these nodes increase by k.
● ADD2 u v k: for edges on the path from u to v, the value of these edges increase by k.
After finished M operation on the tree, please output the value of each node and edge.
Input
The first line of the input is T (1 ≤ T ≤ 20), which stands for the number of test cases you need to solve.
The first line of each case contains two integers N ,M (1 ≤ N, M ≤105),denoting the number of nodes and operations, respectively.
The next N - 1 lines, each lines contains two integers u, v(1 ≤ u, v ≤ N ), denote there is an edge between u,v and its initial value is 0.
For the next M line, contain instructions “ADD1 u v k” or “ADD2 u v k”. (1 ≤ u, v ≤ N, -105 ≤ k ≤ 105)
Output
For each test case, print a line “Case #t:”(without quotes, t means the index of the test case) at the beginning.
The second line contains N integer which means the value of each node.
The third line contains N - 1 integer which means the value of each edge according to the input order.
Sample Input
2 4 2 1 2 2 3 2 4 ADD1 1 4 1 ADD2 3 4 2 4 2 1 2 2 3 1 4 ADD1 1 4 5 ADD2 3 2 4
Sample Output
Case #1: 1 1 0 1 0 2 2 Case #2: 5 0 0 5 0 4 0
题意：两种操作，将u和v路径之间点的权值+w，将u和v路径之间边的权值+w
思路：树链剖分裸题
这题就是类似于在一段区间上+一个数，可以用经典的方法，只需在区间头+w，在区间尾-w即可
先将树形转线性，然后对每条链进行上述操作就好了
对于边操作有一点要注意，因为每个点只与它父亲的边绑定，所以要更新u和v之间的边的时候，u和v的祖先f所绑定的边是不能更新的
因为那条边是不在u和v的路径上的
这题时间卡的紧，输入挂是必须的