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ユーザーが図形を描画できるようにするコードがあります。これには UIBezierPath を使用しています。しかし、たとえば次のように、形状がそれ自体を横切るかどうかを確認する必要があります 。どうすればこれを見つけることができますか?

編集：私はまだこれを解決していません。パス内の行間のすべてのポイントを配列に保存します。次に、配列をループして、交差する線があるかどうかを調べます。しかし、それは機能しません。交差点がないのに交差点があると言うことがあります。

問題はこの方法のどこかにあると思います。

-(BOOL)pathIntersects:(double *)x:(double *)y {
int count = pathPoints.count;
CGPoint p1, p2, p3, p4;
for (int a=0; a<count; a++) {

    //Line 1
    if (a+1<count) {
        p1 = [[pathPoints objectAtIndex:a] CGPointValue];
        p2 = [[pathPoints objectAtIndex:a+1] CGPointValue];
    }else{
        return NO;
    }

    for (int b=0; b<count; b++) {

        //Line 2
        if (b+1<count) {
            p3 = [[pathPoints objectAtIndex:b] CGPointValue];
            p4 = [[pathPoints objectAtIndex:b+1] CGPointValue];
        }else{
            return NO;
        }


        if (!CGPointEqualToPoint(p1, p3) && !CGPointEqualToPoint(p2, p3) && !CGPointEqualToPoint(p4, p1) && !CGPointEqualToPoint(p4, p2)
            && !CGPointEqualToPoint(p1, p2) && !CGPointEqualToPoint(p3, p4)) {

            if (LineIntersect(p1.x, p1.y, p2.x, p2.y, p3.x, p3.y, p4.x, p4.y, x, y)) { 
                return YES;
            }

        }

    }
}
return NO;
}

これは、2 つの線が交差するかどうかを確認するために見つけたコードです。これは C ですが、動作するはずです。

int LineIntersect(
              double x1, double y1,
              double x2, double y2,
              double x3, double y3,
              double x4, double y4,
              double *x, double *y)
{
double mua,mub;
double denom,numera,numerb;

denom  = (y4-y3) * (x2-x1) - (x4-x3) * (y2-y1);
numera = (x4-x3) * (y1-y3) - (y4-y3) * (x1-x3);
numerb = (x2-x1) * (y1-y3) - (y2-y1) * (x1-x3);

/* Are the line coincident? */
if (ABS(numera) < 0.00001 && ABS(numerb) < 0.00001 && ABS(denom) < 0.00001) {
    *x = (x1 + x2) / 2;
    *y = (y1 + y2) / 2;
    return(TRUE);
}

/* Are the line parallel */
if (ABS(denom) < 0.00001) {
    *x = 0;
    *y = 0;
    return(FALSE);
}

/* Is the intersection along the the segments */
mua = numera / denom;
mub = numerb / denom;
if (mua < 0 || mua > 1 || mub < 0 || mub > 1) {
    *x = 0;
    *y = 0;
    return(FALSE);
}
*x = x1 + mua * (x2 - x1);
*y = y1 + mua * (y2 - y1);
return(TRUE);
}
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1 に答える 1

3

It depends on how complex the polygon drawn by the user can be and the number of points in the path. Ideally, there would be a point for all the vertices in the shape and nothing more. Get a CGPath from the UIBezierPath and use GCPathApply to hand the elements to a function, which adds each point to an array. Traverse the array with two for loops, one nested in the other, which checks each line segment against every line segment after it using a standard line-line intersection test. As soon as an intersection has been found, break from the loop. Or, if this were a convenience method, return a BOOL. That's the simplest way.

EDIT: Here's an example of a line-line intersection function which returns a BOOL telling you whether or not two segments cross. Pass in the two points that create the first segment followed by the two points that make the second segment. It was hastily modified from a piece of source code I found online quickly, but it works.

CGPoint lineSegmentsIntersect(CGPoint L1P1, CGPoint L1P2, CGPoint L2P1, CGPoint L2P2) 
{ 
    float x1 = L1P1.x, x2 = L1P2.x, x3 = L2P1.x, x4 = L2P2.x;
    float y1 = L1P1.y, y2 = L1P2.y, y3 = L2P1.y, y4 = L2P2.y;

    float bx = x2 - x1; 
    float by = y2 - y1; 
    float dx = x4 - x3; 
    float dy = y4 - y3;

    float b_dot_d_perp = bx * dy - by * dx;

    if(b_dot_d_perp == 0) {
        return NO;
    }

    float cx = x3 - x1;
    float cy = y3 - y1;
    float t = (cx * dy - cy * dx) / b_dot_d_perp;

    if(t < 0 || t > 1) {
        return NO;
    }

    float u = (cx * by - cy * bx) / b_dot_d_perp;

    if(u < 0 || u > 1) { 
        return NO;
    }

    return YES;
}

You can use it like this.

if (lineSegmentsIntersect(lineOnePointOne,lineOnePointTwo,lineTwoPointOne,lineTwoPointTwo)){
     //segments intersect
} else {
     //segments did not intersect
}

It's up to you to create the double loop to check the correct segments against one another.

于 2012-11-15T17:51:41.277 に答える