树形dp____Magic boy Bi Luo with his excited tree( hdu 5834 2016ccpc网络赛)
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
题意：
一个n个节点的数，每个节点有一个价值为pi的宝藏，点之间的边有权值，每次经过该边会扣掉该边权值对应的价值(价值可以为负数继续前行).问分别从每点开始走，最多可以获得多少价值？
分析：
树形dp。需要讨论的细节比较多。
需要记录两个数据。f[N][2],g[N][2]。以节点1为根节点。
对于f[i][0/1]：f[i][0],表示从i点开始走i - i父亲支路并且回到i点的最大价值，f[i][1] 表示从i开始走i - i父亲支路并且不用回到i点的最大价值。（注意：都不包括i点本身的价值).
对于g[i][0/1]：g[i][0]:表示从i点开始走向子孙节点并且回到i点的最大价值,g[i][1] 表示从i点开始走向子孙节点并且不用回到i点的最大价值。( 注意：都包括i点本身价值).
所以对于任意一点的最大价值为： max( g[i][0]+f[i][1] , g[i][1]+f[i][0] ).
至于f,g的状态转移看代码吧。
代码：
#include