Please Help DFS code explaination

[sk1234]

I want to solve this question as i am new to trees and i could not able to understand the editorial.
https://www.codechef.com/APRIL19B/problems/SUBREM

int dfs(int v = 1, int p = 1) {
int res = 0;
for(auto u: g[v]) {
if(u != p) {
res += dfs(u, v);
}
}
return max(A[v] + res, -x);
}

I could not able to understand how this code works.Please help me to clarify my problem ...

[dhayden]

There isn't enough info here to tell what the code is doing. Can you post the rest of the code?

[sk1234]

Here is the code:

#include <bits/stdc++.h>

using namespace std;

#define all(x) begin(x), end(x)
#define int int64_t

const int maxn = 1e5 + 42;
vector<int> g[maxn];
int x;
int A[maxn];

void init() {
	for(int i = 0; i < maxn; i++) {
		g[i].clear();
	}
}

int dfs(int v = 1, int p = 1) {
	int res = 0;
	for(auto u: g[v]) {
		if(u != p) {
			res += dfs(u, v);
		}
	}
	return max(A[v] + res, -x);
}

void solve() {
	init();
	int n;
	cin >> n >> x;
	for(int i = 1; i <= n; i++) {
		cin >> A[i];
	}
	for(int i = 1; i < n; i++) {
		int x, y;
		cin >> x >> y;
		g[x].push_back(y);
		g[y].push_back(x);
	}
	cout << dfs() << endl;
}

signed main() {
	//freopen("input.txt", "r", stdin);
	//freopen("output.txt", "w", stdout);
	ios::sync_with_stdio(0);
	cin.tie(0);
	int t;
	cin >> t;
	while(t--) {
		solve();
	}
	return 0;
}

[dhayden]

Here's a description of dfs(). I can't say for sure that it solves the problem.

// recursively determine the profit of the tree starting at node v.
// p is the parent node and it's passed in to prevent the function
// from recursing back to the parent.
int dfs(int v = 1, int p = 1) {
        int res = 0;
        for(auto u: g[v]) {     // for each neighbor of node v
                if(u != p) {    // skip the parent
                        res += dfs(u, v); // add up the profits
                }
        }
        // The profit if you keep this node is this node's value
        // (A[v] plus the profit of all children (res).
        // If you delete this node, the profit is the cost of
        // deleting the node (-x).
        return max(A[v] + res, -x);
}