HackerEarth Crazy Matrix problem solution

YASH PAL,
In this HackerEarth Crazy Matrix problem solution Praveen went crazy yesterday and created an arbitrary matrix consisting of 0, 1 or 2. There was no rule observed in forming this matrix. But then he made up some rules himself.
If there are adjacent 1’s, they are said to be connected. Similarly, if there are adjacent 2’s, they are said to be connected. Here adjacent means all the surrounding positions along with the diagonal positions.
Now given a matrix of 0, 1 or 2, do your computation according to the following rules:
  1. If there is a path from any position in top row to any position in bottom row consisting only of 1, then print 1.
  2. If there is a path from any position in first column to any position in last column consisting only of 2, then print 2.
  3. If both Rule 1 & Rule 2 are true, print AMBIGUOUS.
  4. If none of Rule 1, Rule 2 or Rule 3 satisfies, print 0.
HackerEarth Crazy Matrix problem solution

HackerEarth Crazy Matrix problem solution.

#include <bits/stdc++.h>

using namespace std;

int n;
int arr [100 + 10][100 + 10];
int dirc [8][2] = { {0 , 1}, {0 , -1}, {1 , 0}, {-1 , 0}, {1 , 1}, {1 , -1}, {-1 , -1}, {-1 , 1} };     // all the possible directions

bool vis1 [100 + 10][100 + 10];             // for first dfs
bool vis2 [100 + 10][100 + 10];             // for second dfs

bool dfs(int x, int y, int comp, bool vis [][100 + 10]){

    if(comp == 1 && x == n - 1) return true;        // base case in first dfs
    if(comp == 2 && y == n - 1) return true;        // base case in second dfs

    vis[x][y] = true;

    bool ans = false;

    for(int i = 0; i < 8 && !ans; i++){

        int dx = x + dirc[i][0];                    // transition
        int dy = y + dirc[i][1];

        if(dx < 0 || dx >= n || dy < 0|| dy >= n) continue;     // out of the board

        if(arr[dx][dy] == comp && !vis[dx][dy])
            ans |= dfs(dx, dy, comp, vis);          
    }

    return ans;
}

int main()
{
    ios::sync_with_stdio(false);cin.tie(0);             // decrease the time for cin, cout


    cin >>     n           ;

    for(int i = 0; i < n; i++){

        for(int j = 0; j < n; j++){

            cin >> arr[i][j];
        }
    }

    bool ones = false, twos = false;

    for(int i = 0; i < n  && !ones ; i++){

        if(arr[0][i] == 1 && !vis1[0][i]) ones = dfs(0 , i , 1, vis1);      // check for a path from the first row
    }

    for(int i = 0; i < n  && !twos ; i++){

        if(arr[i][0] == 2 && !vis2[i][0]) twos = dfs(i , 0 , 2, vis2);      // check for a path from the second row
    }

    if(ones && twos) cout << "AMBIGUOUS";
    else if (ones) cout << "1";
    else if (twos) cout << "2";
    else cout << "0";

    return 0;
}
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