確定兩個(gè)矩形是否重疊？我正在嘗試編寫(xiě)一個(gè)C+程序，它接受用戶(hù)的以下輸入來(lái)構(gòu)造矩形(介于2到5之間)：高度、寬度、x-pos、y-pos。所有這些矩形都將與x和y軸平行存在，即它們的所有邊都有0或無(wú)窮大的斜率。我試著去實(shí)現(xiàn)這，這個(gè)有疑問(wèn)，但我運(yùn)氣不太好。我目前的實(shí)現(xiàn)如下：// Gets all the vertices for Rectangle 1 and stores them in an array -> arrRect1
// point 1 x: arrRect1[0], point 1 y: arrRect1[1] and so on...
// Gets all the vertices for Rectangle 2 and stores them in an array -> arrRect2
// rotated edge of point a, rect 1int rot_x, rot_y;rot_x = -arrRect1[3];rot_y = arrRect1[2];
// point on rotated edgeint pnt_x, pnt_y;pnt_x = arrRect1[2]; pnt_y = arrRect1[3];
// test point, a from rect 2int tst_x, tst_y;tst_x = arrRect2[0];tst_y = arrRect2[1];
int value;value = (rot_x * (tst_x - pnt_x)) + (rot_y * (tst_y - pnt_y));cout << "Value: " << value;但是，我不太確定(A)我是否正確地實(shí)現(xiàn)了我鏈接到的算法，或者我是否準(zhǔn)確地解釋了這一點(diǎn)？有什么建議嗎？
3 回答
心有法竹
TA貢獻(xiàn)1866條經(jīng)驗(yàn) 獲得超5個(gè)贊
struct rect{
int x;
int y;
int width;
int height;};bool valueInRange(int value, int min, int max){ return (value >= min) && (value <= max); }bool rectOverlap(rect A, rect B){
bool xOverlap = valueInRange(A.x, B.x, B.x + B.width) ||
valueInRange(B.x, A.x, A.x + A.width);
bool yOverlap = valueInRange(A.y, B.y, B.y + B.height) ||
valueInRange(B.y, A.y, A.y + A.height);
return xOverlap && yOverlap;}
慕虎7371278
TA貢獻(xiàn)1802條經(jīng)驗(yàn) 獲得超4個(gè)贊
struct Rect{
Rect(int x1, int x2, int y1, int y2)
: x1(x1), x2(x2), y1(y1), y2(y2)
{
assert(x1 < x2);
assert(y1 < y2);
}
int x1, x2, y1, y2;};booloverlap(const Rect &r1, const Rect &r2){
// The rectangles don't overlap if
// one rectangle's minimum in some dimension
// is greater than the other's maximum in
// that dimension.
bool noOverlap = r1.x1 > r2.x2 ||
r2.x1 > r1.x2 ||
r1.y1 > r2.y2 ||
r2.y1 > r1.y2;
return !noOverlap;}添加回答
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