有人可以解释如何检查一个旋转的矩形是否与其他矩形相交？
答案 0 :(得分：30)
/// Checks if the two polygons are intersecting.
bool IsPolygonsIntersecting(Polygon a, Polygon b)
{
foreach (var polygon in new[] { a, b })
{
for (int i1 = 0; i1 < polygon.Points.Count; i1++)
{
int i2 = (i1 + 1) % polygon.Points.Count;
var p1 = polygon.Points[i1];
var p2 = polygon.Points[i2];
var normal = new Point(p2.Y - p1.Y, p1.X - p2.X);
double? minA = null, maxA = null;
foreach (var p in a.Points)
{
var projected = normal.X * p.X + normal.Y * p.Y;
if (minA == null || projected < minA)
minA = projected;
if (maxA == null || projected > maxA)
maxA = projected;
}
double? minB = null, maxB = null;
foreach (var p in b.Points)
{
var projected = normal.X * p.X + normal.Y * p.Y;
if (minB == null || projected < minB)
minB = projected;
if (maxB == null || projected > maxB)
maxB = projected;
}
if (maxA < minB || maxB < minA)
return false;
}
}
return true;
}
有关详细信息，请参阅此文章：2D Polygon Collision Detection - Code Project
NB：该算法仅适用于凸多边形，以顺时针或逆时针顺序指定。
答案 1 :(得分：29)
在javascript中，完全相同的算法（为方便起见）：
/**
* Helper function to determine whether there is an intersection between the two polygons described
* by the lists of vertices. Uses the Separating Axis Theorem
*
* @param a an array of connected points [{x:, y:}, {x:, y:},...] that form a closed polygon
* @param b an array of connected points [{x:, y:}, {x:, y:},...] that form a closed polygon
* @return true if there is any intersection between the 2 polygons, false otherwise
*/
function doPolygonsIntersect (a, b) {
var polygons = [a, b];
var minA, maxA, projected, i, i1, j, minB, maxB;
for (i = 0; i < polygons.length; i++) {
// for each polygon, look at each edge of the polygon, and determine if it separates
// the two shapes
var polygon = polygons[i];
for (i1 = 0; i1 < polygon.length; i1++) {
// grab 2 vertices to create an edge
var i2 = (i1 + 1) % polygon.length;
var p1 = polygon[i1];
var p2 = polygon[i2];
// find the line perpendicular to this edge
var normal = { x: p2.y - p1.y, y: p1.x - p2.x };
minA = maxA = undefined;
// for each vertex in the first shape, project it onto the line perpendicular to the edge
// and keep track of the min and max of these values
for (j = 0; j < a.length; j++) {
projected = normal.x * a[j].x + normal.y * a[j].y;
if (isUndefined(minA) || projected < minA) {
minA = projected;
}
if (isUndefined(maxA) || projected > maxA) {
maxA = projected;
}
}
// for each vertex in the second shape, project it onto the line perpendicular to the edge
// and keep track of the min and max of these values
minB = maxB = undefined;
for (j = 0; j < b.length; j++) {
projected = normal.x * b[j].x + normal.y * b[j].y;
if (isUndefined(minB) || projected < minB) {
minB = projected;
}
if (isUndefined(maxB) || projected > maxB) {
maxB = projected;
}
}
// if there is no overlap between the projects, the edge we are looking at separates the two
// polygons, and we know there is no overlap
if (maxA < minB || maxB < minA) {
CONSOLE("polygons don't intersect!");
return false;
}
}
}
return true;
};
希望这有助于某人。
答案 2 :(得分：7)
如果有人感兴趣的话，这是Java中的相同算法。
boolean isPolygonsIntersecting(Polygon a, Polygon b)
{
for (int x=0; x<2; x++)
{
Polygon polygon = (x==0) ? a : b;
for (int i1=0; i1<polygon.getPoints().length; i1++)
{
int i2 = (i1 + 1) % polygon.getPoints().length;
Point p1 = polygon.getPoints()[i1];
Point p2 = polygon.getPoints()[i2];
Point normal = new Point(p2.y - p1.y, p1.x - p2.x);
double minA = Double.POSITIVE_INFINITY;
double maxA = Double.NEGATIVE_INFINITY;
for (Point p : a.getPoints())
{
double projected = normal.x * p.x + normal.y * p.y;
if (projected < minA)
minA = projected;
if (projected > maxA)
maxA = projected;
}
double minB = Double.POSITIVE_INFINITY;
double maxB = Double.NEGATIVE_INFINITY;
for (Point p : b.getPoints())
{
double projected = normal.x * p.x + normal.y * p.y;
if (projected < minB)
minB = projected;
if (projected > maxB)
maxB = projected;
}
if (maxA < minB || maxB < minA)
return false;
}
}
return true;
}
答案 3 :(得分：2)
检查Oren Becker设计的方法，以检测旋转矩形的交叉形式：
struct _Vector2D
{
float x, y;
};
// C:center; S: size (w,h); ang: in radians,
// rotate the plane by [-ang] to make the second rectangle axis in C aligned (vertical)
struct _RotRect
{
_Vector2D C;
_Vector2D S;
float ang;
};
调用以下函数将返回两个旋转的矩形是否相交：
// Rotated Rectangles Collision Detection, Oren Becker, 2001
bool check_two_rotated_rects_intersect(_RotRect * rr1, _RotRect * rr2)
{
_Vector2D A, B, // vertices of the rotated rr2
C, // center of rr2
BL, TR; // vertices of rr2 (bottom-left, top-right)
float ang = rr1->ang - rr2->ang, // orientation of rotated rr1
cosa = cos(ang), // precalculated trigonometic -
sina = sin(ang); // - values for repeated use
float t, x, a; // temporary variables for various uses
float dx; // deltaX for linear equations
float ext1, ext2; // min/max vertical values
// move rr2 to make rr1 cannonic
C = rr2->C;
SubVectors2D(&C, &rr1->C);
// rotate rr2 clockwise by rr2->ang to make rr2 axis-aligned
RotateVector2DClockwise(&C, rr2->ang);
// calculate vertices of (moved and axis-aligned := 'ma') rr2
BL = TR = C;
/*SubVectors2D(&BL, &rr2->S);
AddVectors2D(&TR, &rr2->S);*/
//-----------------------------------
BL.x -= rr2->S.x/2; BL.y -= rr2->S.y/2;
TR.x += rr2->S.x/2; TR.y += rr2->S.y/2;
// calculate vertices of (rotated := 'r') rr1
A.x = -(rr1->S.y/2)*sina; B.x = A.x; t = (rr1->S.x/2)*cosa; A.x += t; B.x -= t;
A.y = (rr1->S.y/2)*cosa; B.y = A.y; t = (rr1->S.x/2)*sina; A.y += t; B.y -= t;
//---------------------------------------
//// calculate vertices of (rotated := 'r') rr1
//A.x = -rr1->S.y*sina; B.x = A.x; t = rr1->S.x*cosa; A.x += t; B.x -= t;
//A.y = rr1->S.y*cosa; B.y = A.y; t = rr1->S.x*sina; A.y += t; B.y -= t;
t = sina*cosa;
// verify that A is vertical min/max, B is horizontal min/max
if (t < 0)
{
t = A.x; A.x = B.x; B.x = t;
t = A.y; A.y = B.y; B.y = t;
}
// verify that B is horizontal minimum (leftest-vertex)
if (sina < 0) { B.x = -B.x; B.y = -B.y; }
// if rr2(ma) isn't in the horizontal range of
// colliding with rr1(r), collision is impossible
if (B.x > TR.x || B.x > -BL.x) return 0;
// if rr1(r) is axis-aligned, vertical min/max are easy to get
if (t == 0) {ext1 = A.y; ext2 = -ext1; }
// else, find vertical min/max in the range [BL.x, TR.x]
else
{
x = BL.x-A.x; a = TR.x-A.x;
ext1 = A.y;
// if the first vertical min/max isn't in (BL.x, TR.x), then
// find the vertical min/max on BL.x or on TR.x
if (a*x > 0)
{
dx = A.x;
if (x < 0) { dx -= B.x; ext1 -= B.y; x = a; }
else { dx += B.x; ext1 += B.y; }
ext1 *= x; ext1 /= dx; ext1 += A.y;
}
x = BL.x+A.x; a = TR.x+A.x;
ext2 = -A.y;
// if the second vertical min/max isn't in (BL.x, TR.x), then
// find the local vertical min/max on BL.x or on TR.x
if (a*x > 0)
{
dx = -A.x;
if (x < 0) { dx -= B.x; ext2 -= B.y; x = a; }
else { dx += B.x; ext2 += B.y; }
ext2 *= x; ext2 /= dx; ext2 -= A.y;
}
}
// check whether rr2(ma) is in the vertical range of colliding with rr1(r)
// (for the horizontal range of rr2)
return !((ext1 < BL.y && ext2 < BL.y) ||
(ext1 > TR.y && ext2 > TR.y));
}
inline void AddVectors2D(_Vector2D * v1, _Vector2D * v2)
{
v1->x += v2->x; v1->y += v2->y;
}
inline void SubVectors2D(_Vector2D * v1, _Vector2D * v2)
{
v1->x -= v2->x; v1->y -= v2->y;
}
inline void RotateVector2DClockwise(_Vector2D * v, float ang)
{
float t, cosa = cos(ang), sina = sin(ang);
t = v->x;
v->x = t*cosa + v->y*sina;
v->y = -t*sina + v->y*cosa;
}
答案 4 :(得分：2)
也许它会帮助某人。 PHP中的相同算法：
function isPolygonsIntersecting($a, $b) {
$polygons = array($a, $b);
for ($i = 0; $i < count($polygons); $i++) {
$polygon = $polygons[$i];
for ($i1 = 0; $i1 < count($polygon); $i1++) {
$i2 = ($i1 + 1) % count($polygon);
$p1 = $polygon[$i1];
$p2 = $polygon[$i2];
$normal = array(
"x" => $p2["y"] - $p1["y"],
"y" => $p1["x"] - $p2["x"]
);
$minA = NULL; $maxA = NULL;
for ($j = 0; $j < count($a); $j++) {
$projected = $normal["x"] * $a[$j]["x"] + $normal["y"] * $a[$j]["y"];
if (!isset($minA) || $projected < $minA) {
$minA = $projected;
}
if (!isset($maxA) || $projected > $maxA) {
$maxA = $projected;
}
}
$minB = NULL; $maxB = NULL;
for ($j = 0; $j < count($b); $j++) {
$projected = $normal["x"] * $b[$j]["x"] + $normal["y"] * $b[$j]["y"];
if (!isset($minB) || $projected < $minB) {
$minB = $projected;
}
if (!isset($maxB) || $projected > $maxB) {
$maxB = $projected;
}
}
if ($maxA < $minB || $maxB < $minA) {
return false;
}
}
}
return true;
}
答案 5 :(得分：1)
您也可以使用Rect.IntersectsWith()。
例如，在WPF中，如果你有两个UIElements，使用RenderTransform并放在Canvas上，你想知道它们是否相交，你可以使用类似的东西：
bool IsIntersecting(UIElement element1, UIElement element2)
{
Rect area1 = new Rect(
(double)element1.GetValue(Canvas.TopProperty),
(double)element1.GetValue(Canvas.LeftProperty),
(double)element1.GetValue(Canvas.WidthProperty),
(double)element1.GetValue(Canvas.HeightProperty));
Rect area2 = new Rect(
(double)element2.GetValue(Canvas.TopProperty),
(double)element2.GetValue(Canvas.LeftProperty),
(double)element2.GetValue(Canvas.WidthProperty),
(double)element2.GetValue(Canvas.HeightProperty));
Transform transform1 = element1.RenderTransform as Transform;
Transform transform2 = element2.RenderTransform as Transform;
if (transform1 != null)
{
area1.Transform(transform1.Value);
}
if (transform2 != null)
{
area2.Transform(transform2.Value);
}
return area1.IntersectsWith(area2);
}
答案 6 :(得分：0)
带有切换到（ex）的Type（Java）脚本实现包括＆＃34; Touch＆＃34;情况：
class Position {
private _x: number;
private _y: number;
public constructor(x: number = null, y: number = null) {
this._x = x;
this._y = y;
}
public get x() { return this._x; }
public set x(value: number) { this._x = value; }
public get y() { return this._y; }
public set y(value: number) { this._y = value; }
}
class Polygon {
private _positions: Array<Position>;
public constructor(positions: Array<Position> = null) {
this._positions = positions;
}
public addPosition(position: Position) {
if (!position) {
return;
}
if (!this._positions) {
this._positions = new Array<Position>();
}
this._positions.push(position);
}
public get positions(): ReadonlyArray<Position> { return this._positions; }
/**
* https://stackoverflow.com/a/12414951/468910
*
* Helper function to determine whether there is an intersection between the two polygons described
* by the lists of vertices. Uses the Separating Axis Theorem
*
* @param polygonToCompare a polygon to compare with
* @param allowTouch consider it an intersection when polygons only "touch"
* @return true if there is any intersection between the 2 polygons, false otherwise
*/
public isIntersecting(polygonToCompare: Polygon, allowTouch: boolean = true): boolean {
const polygons: Array<ReadonlyArray<Position>> = [this.positions, polygonToCompare.positions]
const firstPolygonPositions: ReadonlyArray<Position> = polygons[0];
const secondPolygonPositions: ReadonlyArray<Position> = polygons[1];
let minA, maxA, projected, i, i1, j, minB, maxB;
for (i = 0; i < polygons.length; i++) {
// for each polygon, look at each edge of the polygon, and determine if it separates
// the two shapes
const polygon = polygons[i];
for (i1 = 0; i1 < polygon.length; i1++) {
// grab 2 vertices to create an edge
const i2 = (i1 + 1) % polygon.length;
const p1 = polygon[i1];
const p2 = polygon[i2];
// find the line perpendicular to this edge
const normal = {
x: p2.y - p1.y,
y: p1.x - p2.x
};
minA = maxA = undefined;
// for each vertex in the first shape, project it onto the line perpendicular to the edge
// and keep track of the min and max of these values
for (j = 0; j < firstPolygonPositions.length; j++) {
projected = normal.x * firstPolygonPositions[j].x + normal.y * firstPolygonPositions[j].y;
if (!minA || projected < minA || (!allowTouch && projected === minA)) {
minA = projected;
}
if (!maxA || projected > maxA || (!allowTouch && projected === maxA)) {
maxA = projected;
}
}
// for each vertex in the second shape, project it onto the line perpendicular to the edge
// and keep track of the min and max of these values
minB = maxB = undefined;
for (j = 0; j < secondPolygonPositions.length; j++) {
projected = normal.x * secondPolygonPositions[j].x + normal.y * secondPolygonPositions[j].y;
if (!minB || projected < minB || (!allowTouch && projected === minB)) {
minB = projected;
}
if (!maxB || projected > maxB || (!allowTouch && projected === maxB)) {
maxB = projected;
}
}
// if there is no overlap between the projects, the edge we are looking at separates the two
// polygons, and we know there is no overlap
if (maxA < minB || (!allowTouch && maxA === minB) || maxB < minA || (!allowTouch && maxB === minA)) {
return false;
}
}
}
return true;
}
答案 7 :(得分：0)
在love2d框架中内置的Lua实现。碰撞检测功能仍然可以在纯lua中工作
math.inf = 1e309
function love.load()
pol = {{0, 0}, {30, 2}, {8, 30}}
pol2 = {{60, 60}, {90, 61}, {98, 100}, {80, 100}}
end
function love.draw()
for k,v in ipairs(pol) do
love.graphics.line(pol[k][1], pol[k][2], pol[k % #pol + 1][1], pol[k % #pol + 1][2])
end
for k,v in ipairs(pol2) do
love.graphics.line(pol2[k][1], pol2[k][2], pol2[k % #pol2 + 1][1], pol2[k % #pol2 + 1][2])
end
end
function love.update(dt)
pol[1][1] = love.mouse.getX()
pol[1][2] = love.mouse.getY()
pol[2][1] = pol[1][1] + 30
pol[2][2] = pol[1][2] + 2
pol[3][1] = pol[1][1] + 8
pol[3][2] = pol[1][2] + 30
--lazy way to see that's function works
print(doPolygonsIntersect(pol, pol2))
end
-------------------------------------------------------------------------
function doPolygonsIntersect(a,b)
polygons = {a,b}
for i=1, #polygons do
polygon = polygons[i]
for i1=1, #polygon do
i2 = i1 % #polygon + 1
p1 = polygon[i1]
p2 = polygon[i2]
nx,ny = p2[2] - p1[2], p1[1] - p2[1]
minA = math.inf
maxA = -math.inf
for j=1, #a do
projected = nx * a[j][1] + ny * a[j][2]
if projected < minA then minA = projected end
if projected > maxA then maxA = projected end
end
minB = math.inf
maxB = -math.inf
for j=1, #b do
projected = nx * b[j][1] + ny * b[j][2]
if projected < minB then minB = projected end
if projected > maxB then maxB = projected end
end
if maxA < minB or maxB < minA then return false end
end
end
return true
end
答案 8 :(得分：0)
在Python中：
def do_polygons_intersect(a, b):
"""
* Helper function to determine whether there is an intersection between the two polygons described
* by the lists of vertices. Uses the Separating Axis Theorem
*
* @param a an ndarray of connected points [[x_1, y_1], [x_2, y_2],...] that form a closed polygon
* @param b an ndarray of connected points [[x_1, y_1], [x_2, y_2],...] that form a closed polygon
* @return true if there is any intersection between the 2 polygons, false otherwise
"""
polygons = [a, b];
minA, maxA, projected, i, i1, j, minB, maxB = None, None, None, None, None, None, None, None
for i in range(len(polygons)):
# for each polygon, look at each edge of the polygon, and determine if it separates
# the two shapes
polygon = polygons[i];
for i1 in range(len(polygon)):
# grab 2 vertices to create an edge
i2 = (i1 + 1) % len(polygon);
p1 = polygon[i1];
p2 = polygon[i2];
# find the line perpendicular to this edge
normal = { 'x': p2[1] - p1[1], 'y': p1[0] - p2[0] };
minA, maxA = None, None
# for each vertex in the first shape, project it onto the line perpendicular to the edge
# and keep track of the min and max of these values
for j in range(len(a)):
projected = normal['x'] * a[j][0] + normal['y'] * a[j][1];
if (minA is None) or (projected < minA):
minA = projected
if (maxA is None) or (projected > maxA):
maxA = projected
# for each vertex in the second shape, project it onto the line perpendicular to the edge
# and keep track of the min and max of these values
minB, maxB = None, None
for j in range(len(b)):
projected = normal['x'] * b[j][0] + normal['y'] * b[j][1]
if (minB is None) or (projected < minB):
minB = projected
if (maxB is None) or (projected > maxB):
maxB = projected
# if there is no overlap between the projects, the edge we are looking at separates the two
# polygons, and we know there is no overlap
if (maxA < minB) or (maxB < minA):
print("polygons don't intersect!")
return False;
return True
答案 9 :(得分：0)
使用Sri的JavaScript并将其与Phaser 3多边形一起使用。
webdrivermanager答案 10 :(得分：0)
这里在LUA中，希望它会在需要时帮助到人们：
function doPolygonsIntersect(a, b)
local polygons = { a, b };
local minA, maxA, projected, i, i1, j, minB, maxB;
for i = 1, #polygons do
--// for each polygon, look at each edge of the polygon, and determine if it separates
--// the two shapes
local polygon = polygons[i];
for i1 = 0, (#polygon-1) do
--// grab 2 vertices to create an edge
local i2 = (i1 + 1) % (#polygon);
local p1 = polygon[i1+1];
local p2 = polygon[i2+1];
--// find the line perpendicular to this edge
local normal = { x = p2.y - p1.y, y = p1.x - p2.x };
minA = nil;
maxA = nil;
--// for each vertex in the first shape, project it onto the line perpendicular to the edge
--// and keep track of the min and max of these values
for j = 1, #a do
projected = normal.x * a[j].x + normal.y * a[j].y;
if (minA == nil or projected < minA) then
minA = projected;
end
if (maxA == nil or projected > maxA) then
maxA = projected;
end
end
--// for each vertex in the second shape, project it onto the line perpendicular to the edge
--// and keep track of the min and max of these values
minB = nil;
maxB = nil;
for j = 1, #b do
projected = normal.x * b[j].x + normal.y * b[j].y;
if (minB == nil or projected < minB) then
minB = projected;
end
if (maxB == nil or projected > maxB) then
maxB = projected;
end
end
if (maxA < minB or maxB < minA) then
return false;
end
end
end
return true;
end
答案 11 :(得分：0)
Go 中公认的算法。
func isPolygonIntersecting(a *Polygon, b *Polygon) bool {
for _, polygon := range []*Polygon{a, b} {
for i1 := 0; i1 < len(*polygon); i1++ {
i2 := (i1 + 1) % len(*polygon)
p1 := (*polygon)[i1]
p2 := (*polygon)[i2]
normal := pixel.V(p2.Y-p1.Y, p1.X-p2.X)
minA := math.MaxFloat64
maxA := math.MaxFloat64
for _, p := range *a {
projected := normal.X*p.X + normal.Y*p.Y
if minA == math.MaxFloat64 || projected < minA {
minA = projected
}
if maxA == math.MaxFloat64 || projected > maxA {
maxA = projected
}
}
minB := math.MaxFloat64
maxB := math.MaxFloat64
for _, p := range *b {
projected := normal.X*p.X + normal.Y*p.Y
if minB == math.MaxFloat64 || projected < minB {
minB = projected
}
if maxB == math.MaxFloat64 || projected > maxB {
maxB = projected
}
}
if maxA < minB || maxB < minA {
return false
}
}
}
return true
}
type Polygon []struct {
X float64
Y float64
}