Beautiful Solutions

using struct

when to use struct

• when there are questions where limited containers are there and each container has specific needs
• when its difficult to take inputs for a container
• when pair, map are not sufficient and we tart thinking about tuple

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

#define endl '\n'
#define ll long long

struct Rect
{
    int x1, y1, x2, y2;
    // Rect(int x1, int y1, int x2, int y2) : x1{x1}, y1{y1}, x2{x2}, y2{y2} {}
};

int area(Rect rect)
{
    int ans = ((rect.x2 - rect.x1) * (rect.y2 - rect.y1));
    return ans;
}

int inter(Rect b, Rect t)
{
    Rect in;
    int ar;

    // y1
    in.y1 = (t.y1 <= b.y1) ? b.y1 : t.y1;
    // y2
    if (t.y2 >= b.y1 && t.y2 >= b.y2)
    {
        in.y2 = b.y2;
    }
    else if (t.y2 >= b.y1 && t.y2 <= b.y2)
    {
        in.y2 = t.y2;
    }

    // x1
    if (t.x1 >= b.x1 && t.x1 <= b.x2)
    {
        in.x1 = t.x1;
    }
    else if (t.x1 <= b.x1 && t.x1 <= b.x2)
    {
        in.x1 = b.x1;
    }
    // x2
    in.x2 = (t.x2 >= b.x2) ? b.x2 : t.x2;

    ar = area(in);

    return ar;
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    // freopen("billboard.in", "r", stdin);
    // freopen("billboard.out", "w", stdout);

    int ans;
    Rect b1, b2, t;
    cin >> b1.x1 >> b1.y1 >> b1.x2 >> b1.y2;
    cin >> b2.x1 >> b2.y1 >> b2.x2 >> b2.y2;
    cin >> t.x1 >> t.y1 >> t.x2 >> t.y2;

    int ib1 = inter(b1, t);
    int ib2 = inter(b2, t);

    // cerr << "b1 ar: " << area(b1) << endl;
    ans = area(b1) + area(b2) - ib1 - ib2;

    cout << ans;
    return 0;
}