HackerEarth Root paths problem solution

In this HackerEarth Root paths problem solution Alice and Bob play a game. The game starts with a graph that contains n nodes and no edges. You are required to process q queries of two types:

Add a directed edge from u[i] to v[i].

Print x[i] if there is a directed path from vertex 1 to vertex x[i]. Otherwise, print 0.

HackerEarth Root paths problem solution.

#include <iostream>
#include <vector>

using namespace std;

class Solver {
    public:
    
    int n, q, t;
    Solver(int nn) {
        n = nn;
        color.resize(n + 1);
        G.resize(n + 1);
        color[1] = 1;

    }

    vector <int> color;
    vector <vector<int> > G;

    inline void repaint(int x) {
        while (!G[x].empty()) {
            int to = G[x].back();
            G[x].pop_back();

            if (color[to]) {
                continue;
            } else {
                color[to] = 1;
                repaint(to);
            }
        }
    }

    inline void add_edge(int u, int v) {
        if (color[v]) {
            return;
        }
        if (color[u]) {
            color[v] = 1;
            repaint(v);
        } else {
            G[u].push_back(v);
        }
    }

    inline int get_ans(int x) {
        if (color[x]) {
            return x;
        }
        return 0;
    }

};
int main() {

        ios_base::sync_with_stdio(false);
        cin.tie(0);
        int n, q, t;
        cin >> n >> q >> t;

        Solver *s = new Solver(n);
        int lastans = 0;
        for (int i = 1; i <= q; i++) {
            int type;
            cin >> type;
            if (type == 1) {
                int a, b;
                cin >> a >> b;
                a = (a ^ (t * lastans)) % n + 1;
                b = (b ^ (t * lastans)) % n + 1;
                s->add_edge(a, b);
            } else {
                int a;
                cin >> a;
                a = (a ^ (t * lastans)) % n + 1;
                lastans = s->get_ans(a);
                cout << lastans << '\n';
            }
        }

    return 0;
}