Promblem & Approach

The problem asks for the minimum area of a rectangle that contains all but three cows, after removing any three cows from the set. The main challenge is to efficiently determine which three cows to remove to minimize the area.

Approach

• Struct Definiton

  • You define a Point struct to represent each cow’s coordinates and a disable flag to mark cows that are considered for removal.

• Sorting and Marking Candidates

  • You sort the cows by their x and y coordinates.
  • You mark the three cows with the smallest and largest x and y values as candidates for removal, since removing cows from the edges is most likely to reduce the rectangle’s area.

• Generating Combinations

  • You collet all marked cows into a chosen vector
  • You generate all possible combinations of three cows from this set using bitmasking

• Area Calculation

  • For each combination, you remove the selected three cows and calculate the area of the smallest rectangle that contains the remaining cows.
  • You keep track of teh minimum area found

• Output

  • The minimum area is printed as the answer.

Related Concepts

1. Sorting
2. Bitmasking
3. Greedy
4. Brute Force
5. Geometry

This method is efficient because it leverages the geometric property that only cows on the boundary can affect the minimum bounding rectangle, and it uses combinatorial techniques to try all relevant removals.

Implementation with C++

int N;

struct Point {
    int x;
    int y;
    bool disable;
};

vector<Point> cows;
// vector<Point> chosen;

int find(int one, int two, int three) {

    int minX = INT_MAX, maxX = INT_MIN;
    int minY = INT_MAX, maxY = INT_MIN;


    F0R(i, N) {
        if (i != one && i != two && i != three) {
            minX = min(minX, cows[i].x);
            maxX = max(maxX, cows[i].x);
            minY = min(minY, cows[i].y);
            maxY = max(maxY, cows[i].y);
        }
    }

    return (maxX - minX) * (maxY - minY);
}


bool valid(int i) {
    int sum = 0;
    while (i) {
        sum += i & 1;
        i = i >> 1;
    }
    return sum == 3;
}

vector<vector<int>> combinations(int num) {
    vector<vector<int>> ans;

    F0R(i, pow(2, num)) {
        if (valid(i)) {
            //dbg(bitset<12>(i));
            vector<int> x;
            int k = i;
            F0R(j, num) {
                if (k & 1) {
                    x.push_back(j);
                }
                k = k >> 1;
            }
            ans.push_back(x);
        }
    }
    dbg(ans.size());
    return ans;
}


int main() {
    ifstream cin("reduce.in");
    ofstream cout("reduce.out");

    cin >> N;

    cows.resize(N);

    F0R(i, N) {
        cin >> cows[i].x >> cows[i].y;
    }

    sort(cows.begin(), cows.end(), [](Point a, Point b) {return a.x < b.x;});

    cows[0].disable = true;
    cows[1].disable = true;
    cows[2].disable = true;

    cows[N - 1].disable = true;
    cows[N - 2].disable = true;
    cows[N - 3].disable = true;


    sort(cows.begin(), cows.end(), [](Point a, Point b) {return a.y < b.y;});

    cows[0].disable = true;
    cows[1].disable = true;
    cows[2].disable = true;

    cows[N - 1].disable = true;
    cows[N - 2].disable = true;
    cows[N - 3].disable = true;

    vector<int> chosen;

    F0R(i, cows.size()) {
        if (cows[i].disable) {
            chosen.push_back(i);
        }
    }


    dbg(chosen);

    vector<vector<int>> myVec = combinations(chosen.size());

    int ans = INT_MAX;


    for(auto choice : myVec) {
        int area = find(chosen[choice[0]], chosen[choice[1]], chosen[choice[2]]);
        dbg(choice, chosen[choice[0]], chosen[choice[1]], chosen[choice[2]], area);
        ans = min(ans, area);
    }

    cout << ans;
}