Codeforces 1092F

To a tree, for each node u, there is a value of sum (dist (u, v) * val [v]), v is the other nodes, find the maximum value.
Tree dp
If u v is a child node, places v is the root of the subtree contribution is sub-tree and for u all nodes.
U is to first find the root of the subtree and weights. Sum [v], - can be considered while dfs while the answer, when transferred from u to v, the current value to be subtracted sum [v], plus the sum [u]
when the above two steps miss, the key is to calculate v , u is the equivalent of the child node v, the sum [u] - = sum [ v], sum [v] + = sum [u], turn a bit corresponding to the tree.
2e5 * 2e5 will burst int, to be explicit with LL transition

#include <bits/stdc++.h>

using namespace std;
typedef long long LL;
const int N = 2e5 + 10;
int a[N],u,v,n;
vector<int> G[N];
LL tmp, ans,sum[N];
void dfs(int u,int fa,int st){
	tmp += (LL)st * a[u];
	sum[u] = a[u];
	for(int i = 0;i<G[u].size();i++){
		int v = G[u][i];
		if(v == fa) continue;
		dfs(v,u,st+1);
		sum[u] += sum[v];
	}
}
void mov(int u,int fa){
	ans = max(ans,tmp);
	for(int i = 0;i<G[u].size();i++){
		int v = G[u][i];
		if(v == fa) continue;
		tmp -= sum[v];
		tmp += (sum[u] - sum[v]);
		sum[u] -= sum[v];
		sum[v] += sum[u];
		mov(v,u);
		sum[v] -= sum[u];
		sum[u] += sum[v];
		tmp -= (sum[u] - sum[v]);
		tmp += sum[v];
	}
}
int main(){
	ios::sync_with_stdio(false);
	cin.tie(0);cout.tie(0);
	cin>>n;
	for(int i = 1;i<=n;i++) cin>>a[i];
	for(int i = 0;i<n-1;i++){
		cin>>u>>v;
		G[u].push_back(v);
		G[v].push_back(u);
	}
	dfs(1,-1,0);
	mov(1,-1);
	cout<<ans;
	return 0;
}