# HDU 6201 transaction transaction transaction （树形dp）

## Description

Kelukin is a businessman. Every day, he travels around cities to do some business. On August 17th, in memory of a great man, citizens will read a book named "the Man Who Changed China". Of course, Kelukin wouldn't miss this chance to make money, but he doesn't have this book. So he has to choose two city to buy and sell.

As we know, the price of this book was different in each city. It is ai yuan in it city. Kelukin will take taxi, whose price is 1yuan per km and this fare cannot be ignored.

There are n−1 roads connecting n cities. Kelukin can choose any city to start his travel. He want to know the maximum money he can get.

## Input

The first line contains an integer T (1≤T≤10) , the number of test cases.

For each test case:

first line contains an integer n (2≤n≤100000) means the number of cities;

second line contains n numbers, the ith number means the prices in ith city; (1≤Price≤10000)

then follows n−1 lines, each contains three numbers x, y and z which means there exists a road between x and y, the distance is zkm (1≤z≤1000).

## Output

For each test case, output a single number in a line: the maximum money he can get.

1
4
10 40 15 30
1 2 30
1 3 2
3 4 10

8

## 思路

$out[x]$ 代表 $x$ 的子树中找一点卖一本书到当前点的最大利益。

## AC 代码

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
#define inf 0x3f3f3f3f
const int maxn = 2e5+10;
struct node
{
int to;
int next;
int cost;
} edge[maxn];
int n;
int pri[maxn];
LL ans;
LL in[maxn];
LL out[maxn];

void init()
{
memset(in,-inf,sizeof(in));
ans=tot=0;
}

{
edge[tot].to=v;
edge[tot].cost=cost;
}

void dfs(int x,int fa)
{
in[x]=-pri[x];
out[x]=pri[x];
{
int to=edge[i].to;
if(to==fa)continue;
dfs(to,x);
in[x]=max(in[x],in[to]-edge[i].cost);
out[x]=max(out[x],out[to]-edge[i].cost);
ans=max(ans,in[x]+out[x]);
}
}

int main()
{
int T;
ios::sync_with_stdio(false);
cin>>T;
while(T--)
{
init();
cin>>n;
for(int i=1; i<=n; i++)
cin>>pri[i];
for(int i=1; i<n; i++)
{
int u,v,cost;
cin>>u>>v>>cost;