HackerEarth Min-Max Weighted Edge problem solution

1. the sum of the weights of all the edges is equal to S.
2. for every possible diameter of the tree, the maximum weight over all the edges covered by its path is the minimum possible.

#include <bits/stdc++.h>

using namespace std;

const int N = 2E3 + 5;
int len;
int d_len;
int start;
int vis[N];
int path[N];
set<int> nodes;
vector<int> v[N];

void dfs(int src) {
    len ++;
    if(len > d_len) {
        d_len = len;
        start = src;
    }
    vis[src] = 1;
    for(auto i : v[src]) {
        if(!vis[i])
            dfs(i);
    }
    len --;
    return;
}

void dfs_again(int src) {
    len ++;
    vis[src] = 1;
    path[len] = src;
    if(len == d_len) {
        for(int i = 1; i <= len; i ++)
            nodes.insert(path[i]);
    }
    for(auto i : v[src]) {
        if(!vis[i])
            dfs_again(i);
    }
    len --;
    return;
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int t;
    cin >> t;
    while(t --) {
        int n, s;
        cin >> n >> s;
        int x, y;
        for(int i = 1; i < n; i ++) {
            cin >> x >> y;
            v[x].push_back(y);
            v[y].push_back(x);
        }
        
        len = d_len = 0;
        memset(vis, 0, sizeof vis);
        dfs(1);
        len = d_len = 0;
        memset(vis, 0, sizeof vis);
        dfs(start);
        for(int i = 1; i <= n; i ++) {
            len = 0;
            memset(vis, 0, sizeof vis);
            dfs_again(i);
        }
        bool not_fnd = 0;
        for(int i = 1; i <= n; i ++) {
            if(nodes.find(i) == nodes.end()) {
                not_fnd = 1;
                break;
            }
        }
        int ans = 0;
        if(!not_fnd) {
            ans = s / (n - 1);
            if(s % (n - 1) != 0)
                ans ++;
        }
        cout << ans << 'n';
        nodes.clear();
        for(int i = 1; i <= n; i ++)
            v[i].clear();
    }
    return 0;
}