Why am I getting WA for BUGLIFE on spoj

Question

Hi i am doing this problem https://www.spoj.com/problems/BUGLIFE/ on SPOJ but i am getting WA, can anyone help. Here is my code.

I'm trying to use sets to solve this problem. I've heard about using Bipartite Graphs to solve this problem but I think using this approach should suffice unless there's some fault with my approach . I've tried quite a few Testcases , but I don't know where my code is failing .

Additional Test cases for anyone who's willing to help :-
http://spojtoolkit.com/history/BUGLIFE

Expected output for the test cases
:- http://spojtoolkit.com/test/BUGLIFE

Sample Input :-
2
3 3
1 2
2 3
1 3
4 2
1 2
3 4

Sample Output :-
Scenario #1:
Suspicious bugs found!
Scenario #2:
No suspicious bugs found!

My code's output is same as expected output .

#include <bits/stdc++.h>

using namespace std;

int main()
{
    int t;
    cin >> t;
    for(int s = 0 ; s < t ; s++ ){
        int bugs , inter; // Bugs and Interactions
        cin >> bugs >> inter;
        map<int,int> isDiscovered , male , female;
        int bug1 , bug2;
        vector<pair<int , int> > b; //stores pair of interactions
        for(int i = 0 ; i < inter ; i++){
            cin >> bug1 >> bug2;
            b.push_back(make_pair(bug1,bug2));
        }
        sort(b.begin() , b.end());

        bool ans = true;

        for(int i = 0 ; i < b.size() ; i++){
            bug1 = b[i].first;
            bug2 = b[i].second;
            //both not classified
            if(isDiscovered.find(bug1) == isDiscovered.end() && isDiscovered.find(bug2) == isDiscovered.end()){
                isDiscovered[bug1]++;
                isDiscovered[bug2]++;
                male[bug1]++;
                female[bug2]++;
            }
            //one classified
            if(isDiscovered.find(bug1) != isDiscovered.end() || isDiscovered.find(bug2) != isDiscovered.end()){
                if(isDiscovered.find(bug1) == isDiscovered.end()){
                    //bug1 does not exist
                    isDiscovered[bug1]++;
                    if(male.find(bug2) == male.end()){
                        //bug2 is female
                        male[bug1]++;
                    }
                    else{
                        //bug2 is male
                        female[bug1]++;
                    }
                }
                else{
                    //bug2 does not exist
                    isDiscovered[bug2]++;
                    if(male.find(bug1) == male.end()){
                        //bug1 is female
                        male[bug2]++;
                    }else{
                        female[bug2]++;
                    }
                }
            }
            //both classified
            if(isDiscovered.find(bug1) != isDiscovered.end() && isDiscovered.find(bug2) != isDiscovered.end()){
                if(male.find(bug1) != male.end() && male.find(bug2) != male.end()){
                    //both males
                    ans = false;
                }
                else if(female.find(bug1) != female.end() && female.find(bug2) != female.end()){
                    //both females
                    ans = false;
                }
            }
            if(ans == false){
                break;
            }
        }
        cout << "Scenario #" << s+1 << ":" << endl;
        if(ans == false){
            cout << "Suspicious bugs found!" << endl;
        }
        else{
            cout << "No suspicious bugs found!" << endl;
        }

    }
    return 0;
}

Answer 1

In case both bugs are new, you unconditionally say that the first one is a male, and the second one is female. It is not necessarily so. We only may assume that they are of different genders, but there is no grounds to assign the gender yet. Try

4 2
2 1
3 4

(same as case 2, with the second pair swapped).