CodeForces - 1084D The Fair Nut and the Best Path 树形dp
The Fair Nut is going to travel to the Tree Country, in which there are n cities. Most of the land of this country is covered by forest. Furthermore, the local road system forms a tree (connected graph without cycles). Nut wants to rent a car in the city u and go by a simple path to city v
. He hasn't determined his way, so it's time to do it.
A filling station is located in every city. Because of strange law, Nut can buy only wi
liters of gasoline in the i
-th city. We can assume, that he has infinite money. Each road has a length, and as soon as Nut drives through this road, the amount of gasoline decreases by length. Of course, Nut can't choose a path, which consists of roads, where he runs out of gasoline. He can buy gasoline in every visited city, even in the first and the last.
A path can consist of only one vertex.
He also wants to find the maximum amount of gasoline that he can have at the end of the path. Help him: count it.
Input
The first line contains a single integer n
(1≤n≤3⋅105
) — the number of cities.
The second line contains n
integers w1,w2,…,wn (0≤wi≤109
) — the maximum amounts of liters of gasoline that Nut can buy in cities.
Each of the next n−1
lines describes road and contains three integers u, v, c (1≤u,v≤n, 1≤c≤109, u≠v), where u and v — cities that are connected by this road and c
— its length.
It is guaranteed that graph of road connectivity is a tree.
Output
Print one number — the maximum amount of gasoline that he can have at the end of the path.
Examples
Input
3
1 3 3
1 2 2
1 3 2
Output
3
Input
5
6 3 2 5 0
1 2 10
2 3 3
2 4 1
1 5 1
Output
7
题意：到达一个点的到这个点的价值，经过一个边花费这个边的价值，求得到的最大价值
题解：dp[i] 保存从i 节点开始走向子节点得到的最大价值，每走一个子代更新一下 结果 和 dp[i] 即可
#include
using namespace std;
typedef long long ll;
const int N=3e5+10;
#define pb push_back
#define mp make_pair
vector > v[N];
int val[N],n;
ll dp[N];
ll ans;
void dfs(int u,int fa)
{
dp[u]=val[u];
ans=max(ans,dp[u]);
for(int i=0;i .