HackerEarth Finding pairs problem solution

In this HackerEarth Finding pairs problem solution You are given a rooted tree with N nodes and node 1 as a root. There is a unique path between any two nodes. Here, d(i,j) is defined as a number of edges in a unique path between nodes i and j.

You have to find the number of pairs (i,j) such that and d(i,j) = d(i,1) – d(j,1).

HackerEarth Finding pairs problem solution.

#include<bits/stdc++.h>
using namespace std;

#define FAST ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
#define ordered_set tree<int, null_type,lessl<int>, rb_tree_tag,tree_order_statistics_node_update>
#define ll long long
#define int long long
#define double long double

#define all(a)      (a).begin(),(a).end()
#define sz(x)       (int)x.size()

#define ff first
#define ss second
#define mp make_pair

#define pb push_back
#define endl "n"

#define f(i,l,r) for(int i=l;i<=r;i++)
#define rf(i,r,l) for(int i=r;i>=l;i--)

#define bp __builtin_popcountll
#define inf 1e18

const int N=1e5+5;
const int M=1e9+7;
vector<int> v[N];
vector<int> lvl(N,0);
void dfs(int x,int p)
{
    for(auto X:v[x])
    {
        if(X==p) continue;
        lvl[X]=lvl[x]+1;
        dfs(X,x);
    }
}
void solve()
{
    int n;
    cin>>n;
    for(int i=1;i<n;i++)
    {
        int t1,t2;
        cin>>t1>>t2;
        v[t1].pb(t2);
        v[t2].pb(t1);
    }
    lvl[1]=1;
    dfs(1,0);
    int ans=0;
    for(int i=1;i<=n;i++)
        ans+=lvl[i];

    cout<<ans<<endl;
}

signed main()
{
    FAST
    
    int t=1;
    // cin>>t;
    for(int tc=1;tc<=t;tc++)
    {
        // cout<<"Case #"<<tc<<": ";
        solve();
    }
}

Second solution

#include<bits/stdc++.h>
using namespace std;

#define FAST ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
#define ordered_set tree<int, null_type,lessl<int>, rb_tree_tag,tree_order_statistics_node_update>
#define ll long long
#define int long long
#define double long double

#define all(a)      (a).begin(),(a).end()
#define sz(x)       (int)x.size()

#define ff first
#define ss second
#define mp make_pair

#define pb push_back
#define endl "n"

#define f(i,l,r) for(int i=l;i<=r;i++)
#define rf(i,r,l) for(int i=r;i>=l;i--)

#define bp __builtin_popcountll
#define inf 1e18

const int N=1e5+5;
const int M=1e9+7;
vector<int> v[N];
vector<int> lvl(N,0);
void dfs(int x,int p)
{
    for(auto X:v[x])
    {
        if(X==p) continue;
        lvl[X]=lvl[x]+1;
        dfs(X,x);
    }
}
void solve()
{
    int n;
    cin>>n;
    for(int i=1;i<n;i++)
    {
        int t1,t2;
        cin>>t1>>t2;
        v[t1].pb(t2);
        v[t2].pb(t1);
    }
    lvl[1]=1;
    dfs(1,0);
    int ans=0;
    for(int i=1;i<=n;i++)
        ans+=lvl[i];

    cout<<ans<<endl;


}

signed main()
{
    FAST
    
    int t=1;
    // cin>>t;
    for(int tc=1;tc<=t;tc++)
    {
        // cout<<"Case #"<<tc<<": ";
        solve();
    }
}

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