HDU 5834 Magic boy Bi Luo with his excited tree
Problem Description
Bi Luo is a magic boy, he also has a migic tree, the tree has
N nodes , in each node , there is a treasure, it's value is
V[i], and for each edge, there is a cost
C[i], which means every time you pass the edge
i , you need to pay
C[i].
You may attention that every V[i] can be taken only once, but for some C[i] , you may cost severial times.
Now, Bi Luo define ans[i] as the most value can Bi Luo gets if Bi Luo starts at node i.
Bi Luo is also an excited boy, now he wants to know every ans[i], can you help him?
You may attention that every V[i] can be taken only once, but for some C[i] , you may cost severial times.
Now, Bi Luo define ans[i] as the most value can Bi Luo gets if Bi Luo starts at node i.
Bi Luo is also an excited boy, now he wants to know every ans[i], can you help him?
Input
First line is a positive integer
T(T≤104) , represents there are
T test cases.
Four each test：
The first line contain an integer N (N≤105).
The next line contains N integers V[i], which means the treasure’s value of node i(1≤V[i]≤104).
For the next N−1 lines, each contains three integers u,v,c , which means node u and node v are connected by an edge, it's cost is c(1≤c≤104).
You can assume that the sum of N will not exceed 106.
Four each test：
The first line contain an integer N (N≤105).
The next line contains N integers V[i], which means the treasure’s value of node i(1≤V[i]≤104).
For the next N−1 lines, each contains three integers u,v,c , which means node u and node v are connected by an edge, it's cost is c(1≤c≤104).
You can assume that the sum of N will not exceed 106.
Output
For the i-th test case , first output Case #i: in a single line , then output
N lines , for the i-th line , output
ans[i] in a single line.
Sample Input
1 5 4 1 7 7 7 1 2 6 1 3 1 2 4 8 3 5 2
Sample Output
Case #1: 15 10 14 9 15
一道颇为恶心的树形dp，要考虑的细节挺多的，想了好久死了一堆脑细胞。
首先，理解题意以后，我们可以知道，如果要计算从某一个点出发的最优值，可以进行一次dfs
以1作为根节点为例，图中我们可以先走1-2在回来2-1在去1-3再到3-5，这样可以得到1 2 3 5四个点的价值减去中间经过的边
可以看出一点，从1出发经过很多的分支，最后一条分支是不需要再回来的。
所以进行树形dp，需要记录三个值，从这个节点向下每次都回来的最优值g[x]，
从这个节点向下最后一次不会来的最优值和次优值dp[x][0]和dp[x][1],顺便记录不会来的是哪一条边f
这样，第一次以1为根的dp可以求出dp[1][0]为1节点的答案，接下来通过这个推导下面相邻节点的答案。
对于x的某个孩子来说，有两种情况，
一种是它向下走不回来，显然这在这在之前的dp中可以得到。
另一种是向上走，这就和x节点的最优值选了那条边有关，如果是选择当前边，那么加上之前的次优值，
不然就加上之前的最优值，当然要先和0比一下，因为可能不走过去。
这样就能推出全部的答案了。
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