9
我正致力于在2D游戏中使用Separting Axis定理的实现。它有点作品,但只是一种。分离轴定理让我疯狂!
我使用这样的:
bool penetration = sat(c1, c2) && sat(c2, c1);
凡c1
和c2
是Convex
类型的,定义为:
class Convex
{
public:
float tx, ty;
public:
std::vector<Point> p;
void translate(float x, float y) {
tx = x;
ty = y;
}
};
(Point
是float x
的结构,float y
)
点按顺时针方向键入。
我当前的代码(忽略Qt的调试):
bool sat(Convex c1, Convex c2, QPainter *debug)
{
//Debug
QColor col[] = {QColor(255, 0, 0), QColor(0, 255, 0), QColor(0, 0, 255), QColor(0, 0, 0)};
bool ret = true;
int c1_faces = c1.p.size();
int c2_faces = c2.p.size();
//For every face in c1
for(int i = 0; i < c1_faces; i++)
{
//Grab a face (face x, face y)
float fx = c1.p[i].x - c1.p[(i + 1) % c1_faces].x;
float fy = c1.p[i].y - c1.p[(i + 1) % c1_faces].y;
//Create a perpendicular axis to project on (axis x, axis y)
float ax = -fy, ay = fx;
//Normalize the axis
float len_v = sqrt(ax * ax + ay * ay);
ax /= len_v;
ay /= len_v;
//Debug graphics (ignore)
debug->setPen(col[i]);
//Draw the face
debug->drawLine(QLineF(c1.tx + c1.p[i].x, c1.ty + c1.p[i].y, c1.p[(i + 1) % c1_faces].x + c1.tx, c1.p[(i + 1) % c1_faces].y + c1.ty));
//Draw the axis
debug->save();
debug->translate(c1.p[i].x, c1.p[i].y);
debug->drawLine(QLineF(c1.tx, c1.ty, ax * 100 + c1.tx, ay * 100 + c1.ty));
debug->drawEllipse(QPointF(ax * 100 + c1.tx, ay * 100 + c1.ty), 10, 10);
debug->restore();
//Carve out the min and max values
float c1_min = FLT_MAX, c1_max = FLT_MIN;
float c2_min = FLT_MAX, c2_max = FLT_MIN;
//Project every point in c1 on the axis and store min and max
for(int j = 0; j < c1_faces; j++)
{
float c1_proj = (ax * (c1.p[j].x + c1.tx) + ay * (c1.p[j].y + c1.ty))/(ax * ax + ay * ay);
c1_min = min(c1_proj, c1_min);
c1_max = max(c1_proj, c1_max);
}
//Project every point in c2 on the axis and store min and max
for(int j = 0; j < c2_faces; j++)
{
float c2_proj = (ax * (c2.p[j].x + c2.tx) + ay * (c2.p[j].y + c2.ty))/(ax * ax + ay * ay);
c2_min = min(c2_proj, c2_min);
c2_max = max(c2_proj, c2_max);
}
//Return if the projections do not overlap
if(!(c1_max >= c2_min && c1_min <= c2_max))
ret = false; //return false;
}
return ret; //return true;
}
我到底做错了什么?它完美地登记碰撞,但在上一个边缘敏感(在我的测试使用三角形和菱形):
//Triangle
push_back(Point(0, -150));
push_back(Point(0, 50));
push_back(Point(-100, 100));
//Diamond
push_back(Point(0, -100));
push_back(Point(100, 0));
push_back(Point(0, 100));
push_back(Point(-100, 0));
我在这个得到这个巨型多动症,请帮我:)
http://u8999827.fsdata.se/sat.png
使用编辑器中的“101”按钮格式化您的文章中的代码通常会大大增加人们回答您的问题的可能性。 – axel22 2010-12-07 10:59:42
我做了格式化,并发布了一个问题图片的链接 – Alex 2010-12-07 11:20:01