백준 16135번 : OBB(Oriented bounding box) / C++

16135번: OBB(Oriented bounding box) (acmicpc.net)

16135번: OBB(Oriented bounding box)

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정말 오랜만에 백준문제를 해결했다 OBB 충돌문제로 개념에대해서 이해하면 쉽게 해결? 가능하다

#include <iostream>
#include <math.h>
#include <vector>
using namespace std;
 
struct Vector2
{
    float x;
    float y;
 
    Vector2 Nomalize()
    {
        float squarSum = sqrtf(x * x + y * y);
        if (squarSum == 1.f)
        {
            return *this;
        }
        else if (squarSum == 0.f)
        {
            return Vector2{ 0.f,0.f };
        }
        float invLength = 1 / sqrtf(squarSum);
 
        return Vector2{ x * invLength, y * invLength };
    }
    
    float Dot(Vector2 _other)
    {
        return x * _other.x + y * _other.y;
    }
};
 
int main()
{
    ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
 
    Vector2 a[4];
    Vector2 b[4];
 
    for (int i = 0; i < 4; ++i)
    {
        cin >> a[i].x >> a[i].y;
    }
    for (int i = 0; i < 4; ++i)
    {
        cin >> b[i].x >> b[i].y;
    }
 
    // 사각형 중심좌표 계산
    Vector2 middleA = { (a[0].x + a[2].x) * 0.5f, (a[0].y + a[2].y) * 0.5f, };
    Vector2 middleB = { (b[0].x + b[2].x) * 0.5f, (b[0].y + b[2].y) * 0.5f, };
 
    Vector2 heightA{ (a[0].x - a[1].x) * 0.5f, (a[0].y - a[1].y) * 0.5f };
    Vector2 widthA{ (a[1].x - a[2].x) * 0.5f, (a[1].y - a[2].y) * 0.5f };
    Vector2 heightB{ (b[0].x - b[1].x) * 0.5f, (b[0].y - b[1].y) * 0.5f };
    Vector2 widthB{ (b[1].x - b[2].x) * 0.5f, (b[1].y - b[2].y) * 0.5f };
 
    Vector2 Distance{ middleB.x - middleA.x, middleB.y - middleA.y };
 
    vector<Vector2> basis{ heightA,widthA,heightB,widthB };
 
    for (int i = 0; i < 4; ++i)
    {
        Vector2 u = basis[i].Nomalize();
        
        float sum = 0.f;
        for (int j = 0; j < 4; ++j)
        {
            sum += abs(u.Dot(basis[j]));
        }
 
        float distance = abs(u.Dot(Distance));
 
        if (distance >= sum)
        {
            cout << "0";
            return 0;
        }
    }
 
    cout << "1";
 
 
    return 0;
}