HackerEarth Strange Road System problem solution

In this HackerEarth Strange Road System problem solution, Marichka will visit HackerLand, the country with N (1 <= N <= 10^5) cities, on her spring holidays. She feels very excited because of this.

During the next Q days one of the two following events happens.

– 1 U V P (1 <= U <= N, 1 <= V <= N, 1 <= P <= 10^9) — hackers build a new bidirectional road with happiness points $P$ between cities U and V

– 2 U V (1 <= U <= N, 1 <= V <= N) — Your task is to answer the query. You are given cities numbers — U and V, and your task is to find maximum happiness Marichka can get after travelling in some way(maybe through some intermediate cities) between cities U and V. If there is no way to get from the city U to the city V between them, then simply output -1.

If Marichka’s happiness before traversing some road was X and road’s happiness points are Y, then after traversing it Marichka’s happiness will be equal to X xor Y. Marichka can traverse one road many times. She also can visit some city many times during her travel.

Initially, there are no roads in HackerLand.

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HackerEarth Strange Road System problem solution.

#include <bits/stdc++.h>
using namespace std;
#define pb push_back
int n , m;
int p[100005]; //Parent array for DSU
int d[100005]; //Distance from the root in DSU
vector<int> b[100005]; //Linear basises for connected components
void add(int x , vector<int> &b) //Function to add x in linear basis
{
    for(int i = 0; i < b.size(); i++) 
    {
        int t= b[i]; 
        if((x ^ t) < x) 
        {
            x ^= t;
        }
    }
    if(x == 0)
        return;
    for(int i = 0; i < b.size(); i++)
    {
        if((x ^ b[i]) < b[i])
            b[i] ^= x;
    }
    b.pb(x); //Add x to linear basis
    sort(b.begin() , b.end() , greater<int>());
}
int mx(int x , vector<int> &b)
{
    for(int i = 0; i < b.size(); i++)
    {
        int t = b[i];
        if((x ^ t) > x)
        {
            x ^= t;
        }
    }
    return x;
}
int find(int u)
{
    if(u == p[u])
        return u;
    int P = p[u]; 
    int root = find(P); 
    
    d[u] ^= d[P]; 
    p[u] = root; 
    return root; 
}
void init(int n) 
{
    for(int i = 0; i < n; i++)
    {
        p[i] = i;
        b[i].clear();
        d[i] = 0;
    }
}
void merge(int u , int v , int w) //Function to merge two vertices in DSU
{
    int pu = find(u) , pv = find(v);
    if(pu == pv) //If they are in the same set 
    {
        add(w ^ d[u] ^ d[v] , b[pu]); 
        return;
    }
    p[pv] = pu; 
    d[pv] = d[v] ^ d[u] ^ w;
    for(auto x : b[pv])
    {
        add(x , b[pu]); //Merging U-basis and V-basis
    }
}
int main()
{
    ios_base::sync_with_stdio(0);
    
    
    int n;
    assert(cin >> n);
    
    assert(1 <= n && n <= 100000);
    init(n);
    int q;
    cin >> q;
    
    assert(1 <= q && q <= 100000);
    while(q--)
    {
        int type;
        assert(cin >> type);
        assert(1 <= type && type <= 2);
        if(type == 1)
        {
            int u , v , w;
            assert(cin >> u >> v >> w);
            assert(1 <= u && u <= n);
            assert(1 <= v && v <= n);
            assert(1 <= w && w <= 1e9);
            u--;v--;
            merge(u , v , w);
        }
        else
        {
            int u , v;
            assert(cin >> u >> v);
            assert(1 <= u && u <= n);
            assert(1 <= v && v <= n);
            u--;v--;
            if(find(u) != find(v))
            {
                cout << -1 << endl;
                continue;
            }
            //cout << d[u] << " " << d[v] << endl;
            
            cout << mx(d[u] ^ d[v] , b[find(u)]) << endl;
        }
    }

}

Second solution

#include <bits/stdc++.h>
using namespace std;
#define pb push_back
int n , m;
int p[100005]; //Parent array for DSU
int d[100005]; //Distance from the root in DSU
vector<int> b[100005]; //Linear basises for connected components
void add(int x , vector<int> &b) //Function to add x in linear basis
{
    for(int i = 0; i < b.size(); i++) // Iterating through all integers in basis to combine x with some elements of linear basis.
    //As the result x should have minimum possible value
    {
        int t= b[i]; 
        if((x ^ t) < x) 
        {
            x ^= t;
        }
    }
    if(x == 0) //If x is equal to zero, there is no need to add x to linear basis
        return;
    for(int i = 0; i < b.size(); i++) //If can make some elements of linear basis smaller, we should do this
    {
        if((x ^ b[i]) < b[i])
            b[i] ^= x;
    }
    b.pb(x); //Add x to linear basis
    sort(b.begin() , b.end() , greater<int>());
}
int mx(int x , vector<int> &b)
{
    for(int i = 0; i < b.size(); i++)
    {
        int t = b[i];
        if((x ^ t) > x)
        {
            x ^= t;
        }
    }
    return x;
}
int find(int u)
{
    if(u == p[u])
        return u;
    int P = p[u]; 
    int root = find(P);
    
    d[u] ^= d[P]; //Keeping d up-to-date
    p[u] = root; //And p
    return root; 
}
void init(int n) // A standart function to initialize DSU
{
    for(int i = 0; i < n; i++)
    {
        p[i] = i;
        b[i].clear();
        d[i] = 0;
    }
}
void merge(int u , int v , int w)
{
    int pu = find(u) , pv = find(v);
    if(pu == pv) //If they are in the same set 
    {
        add(w ^ d[u] ^ d[v] , b[pu]); 
        return;
    }
    p[pv] = pu; 
    d[pv] = d[v] ^ d[u] ^ w; 
    for(auto x : b[pv])
    {
        add(x , b[pu]); //Merging U-basis and V-basis
    }
}
int main()
{
    ios_base::sync_with_stdio(0);
    
    
    int n;
    assert(cin >> n);
    
    assert(1 <= n && n <= 100000);
    init(n);
    int q;
    cin >> q;
    
    assert(1 <= q && q <= 100000);
    while(q--)
    {
        int type;
        assert(cin >> type);
        assert(1 <= type && type <= 2);
        if(type == 1)
        {
            int u , v , w;
            assert(cin >> u >> v >> w);
            assert(1 <= u && u <= n);
            assert(1 <= v && v <= n);
            assert(1 <= w && w <= 1e9);
            u--;v--;
            merge(u , v , w);
        }
        else
        {
            int u , v;
            assert(cin >> u >> v);
            assert(1 <= u && u <= n);
            assert(1 <= v && v <= n);
            u--;v--;
            if(find(u) != find(v))
            {
                cout << -1 << endl;
                continue;
            }
            
            cout << mx(d[u] ^ d[v] , b[find(u)]) << endl;
        }
    }

}

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