USACO 2016 February Contest, Bronze Problem 3. Load Balancing

Problems and Contests
#1

Hi. My code for this problem passes all test cases except for #7 (Run time or memory limit exeeded). However, test case 10 has much bigger data (n=100, b=1000000) than test case 7 (n=100, b=5000) and my code passed. Why is this happening? I’d appreciate any help here.

#include <bits/stdc++.h>
using namespace std;
int WIDTH, HEIGHT;
int TopLeft(int a, int b, vector<vector<bool>> matrix)
{
    int NumberOfCows = 0;
    for (int i = 0; i < b; i++)
        for (int j = 0; j < a; j++)
            NumberOfCows += matrix[i][j];

    return NumberOfCows;
}
int TopRight(int a, int b, vector<vector<bool>> matrix)
{
    int NumberOfCows = 0;
    for (int i = 0; i < b; i++)
        for (int j = a; j < WIDTH; j++)
            NumberOfCows += matrix[i][j];

    return NumberOfCows;
}
int BottomLeft(int a, int b, vector<vector<bool>> matrix)
{
    int NumberOfCows = 0;
    for (int i = b; i < HEIGHT; i++)
        for (int j = 0; j < a; j++)
            NumberOfCows += matrix[i][j];

    return NumberOfCows;
}
int BottomRight(int a, int b, vector<vector<bool>> matrix)
{
    int NumberOfCows = 0;
    for (int i = b; i < HEIGHT; i++)
        for (int j = a; j < WIDTH; j++)
            NumberOfCows += matrix[i][j];

    return NumberOfCows;
}
int main()
{
    freopen("balancing.in","r",stdin);
    freopen("balancing.out","w",stdout);
    ios::sync_with_stdio(0);
    cin.tie(0);
    int n, b;
    cin >> n >> b;
    vector<pair<int, int>> cows(n);
    for (int i = 0; i < n; i++)
        cin >> cows[i].first >> cows[i].second;

    // find width & height of the matrix
    set<int> distinct_x_values;
    set<int> distinct_y_values;
    for (int i = 0; i < n; i++)
    {
        distinct_x_values.insert(cows[i].first);
        distinct_y_values.insert(cows[i].second);
    }
    WIDTH = distinct_x_values.size();
    HEIGHT = distinct_y_values.size();

    // create matrix
    vector<vector<bool>> matrix(HEIGHT, vector<bool>(WIDTH, false));
    for (int i = 0; i < n; i++)
    {
        int x_values_left = 0;
        int y_values_up = 0;
        for (auto x : distinct_x_values)
        {
            if (x < cows[i].first)
                ++x_values_left;
            else
                break;
        }
        for (auto y : distinct_y_values)
        {
            if (y > cows[i].second)
                ++y_values_up;
        }
        matrix[y_values_up][x_values_left] = true;
    }

    // go through all possible pairs of (a, b)
    int ans = 1e9;
    for (int i = 1; i < HEIGHT; i++)
        for (int j = 1; j < WIDTH; j++)
            ans = min(ans, (max(TopLeft(i, j, matrix), (max(TopRight(i, j, matrix), (max(BottomLeft(i, j, matrix), BottomRight(i, j, matrix))))))));

    cout << ans;
    return 0;
}
#2