In this HackerEarth Largest cycle in a tree problem solution, You are given a tree of N nodes and N - 1 edge. Now you need to select two nodes a and b in the tree such that the cycle that will be formed after adding an edge between the two nodes a and b, its length should be maximum. If there is more than one possible answer, you can output any of them.


HackerEarth Largest cycle in a tree problem solution


HackerEarth Largest cycle in a tree problem solution.

#include<bits/stdc++.h>
#define LL long long int
#define M 1000000007
#define endl "\n"
#define eps 0.00000001
LL pow(LL a,LL b,LL m){LL x=1,y=a;while(b > 0){if(b%2 == 1){x=(x*y);if(x>m) x%=m;}y = (y*y);if(y>m) y%=m;b /= 2;}return x%m;}
LL gcd(LL a,LL b){if(b==0) return a; else return gcd(b,a%b);}
LL gen(LL start,LL end){LL diff = end-start;LL temp = rand()%start;return temp+diff;}
using namespace std;
bool visit[1000001];
int dist[100001];
vector<int> graph[1000001];
void dfs(int node){
visit[node] = 1;
for(int i: graph[node]){
if(visit[i] == 0){
visit[i] = 1;
dist[i] = dist[node] + 1;
dfs(i);
}
}
}
int main()
{
ios_base::sync_with_stdio(0);
int n;
assert(cin >> n);
for(int i = 1; i < n; i++){
int u , v;
assert(cin >> u >> v);
graph[u].push_back(v);
graph[v].push_back(u);
}

int a , b;
assert(cin >> a >> b);
cout << 100;
return 0;
assert(a >= 1 && a <= n);
assert(a != b);
assert(b >= 1 && b <= n);
dfs(a);
if(dist[b] == ans){
cout << 10;
}
else{
assert(1 == 0);
}
}

Second solution

#include <bits/stdc++.h>

using namespace std;

const int N = 1E5 + 5;
bool vis[N];
vector<int> tree[N];

int start = -1;
int total = 0;

void dfs(int src, int len) {
vis[src] = true;
if(len > total) {
total = len;
start = src;
}
for(auto i : tree[src]) {
if(!vis[i])
dfs(i, len + 1);
}
return;
}

int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int n;
cin >> n;
int a, b;
for(int i = 0; i < n; i ++) {
cin >> a >> b;
tree[a].push_back(b);
tree[b].push_back(a);
}
dfs(1, 1);
cout << start << ' ';
total = 0;
memset(vis, false, sizeof vis);
dfs(start, 1);
cout << start;
return 0;
}