Solution for the grid of many xor using prim algorithm

gridOfManyXor.cpp

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

struct Node {
    int w;
    pair<int, int> c;
    Node(int wi, int xi, int yi) 
    : w(wi), c(xi, yi)
    {}
    bool operator< (const Node& another) const {
        return w < another.w;
    }
};

long long prim(vector< vector<int> >& mat) {
    set< Node > q; 
    q.insert( Node(0, 0, 0) );
    vector< vector<int> > keys(mat.size(), vector<int>(mat[0].size(), numeric_limits<int>::max()));
    vector< vector<bool> > visited(mat.size(), vector<bool>(mat[0].size(), false));
    keys[0][0] = 0;
    while (!q.empty()) {
        auto u = *q.begin();
        visited[u.c.first][u.c.second] = true;
        q.erase(q.begin());
        for (int i = 0; i < 2; ++i)
            for (int j = 0; j < 2; ++j) {
                int s = 1 - 2*i;
                int x = s * j;
                int y = s * (!j);
                if (u.c.first+x >= 0 && u.c.second+y >=0 && 
                    u.c.first+x < mat.size() && u.c.second+y < mat[0].size()) {
                        
                    int cost = mat[u.c.first+x][u.c.second+y] ^ mat[u.c.first][u.c.second];
                    if (!visited[u.c.first+x][u.c.second+y] && 
                        keys[u.c.first+x][u.c.second+y] > cost) {
                        q.erase( Node(keys[u.c.first+x][u.c.second+y], u.c.first+x, u.c.second+y) );
                        keys[u.c.first+x][u.c.second+y] = cost;
                        q.insert( Node(keys[u.c.first+x][u.c.second+y], u.c.first+x, u.c.second+y) );
                    }
                }
            }
    }
    long long res = 0;
    for (int i = 0; i < mat.size(); ++i) {
        for (int j = 0; j < mat[0].size(); ++j) {
            res += keys[i][j];
        }
    }
    return res;
}

int main()
{
    int T;
    cin >> T;
    for (int t = 0; t < T; ++t) {
        int n, m;
        cin >> n >> m;
        int r1,c1,r2,c2;
        cin >> r1 >> c1 >> r2 >> c2;
        vector< vector<int> > mat(n, vector<int>(m, 0));
        for (int i = 0; i < n; ++i)
            for (int j = 0; j < m; ++j)
                cin >> mat[i][j];
        cout << prim(mat) << "\n";;
    }
    
    return 0;
}