由网友(说好了做朋友的./)分享简介：对于比赛我开发我需要一个算法，可以计算交叉点。我已经解决了这个问题，但我已经做它的方式是非常讨厌的，我希望有人在这里可能有一个更好的解决方案。For a game I am developing I need an algorithm that can calculate intersections. I have...

对于比赛我开发我需要一个算法，可以计算交叉点。我已经解决了这个问题，但我已经做它的方式是非常讨厌的，我希望有人在这里可能有一个更好的解决方案。

For a game I am developing I need an algorithm that can calculate intersections. I have solved the problem, but the way I have done it is really nasty and I am hoping someone here might have a more elegant solution.

一对重present它们之间的连线的终点点。鉴于两对点，做画线相交，如果是这样，在什么时候？

A pair of points represent the end points of a line drawn between them. Given two pairs of points, do the drawn lines intersect, and if so, at what point?

因此​​，例如调用线（AX，AY） - （BX，BY）和（CX，CY） - （DX，DY）

So for example call the lines (A.x, A.y)-(B.x, B.y) and (C.x, C.y)-(D.x, D.y)

谁能想到了解决办法？在任何语言中的解决方案就可以了。

Can anyone think of a solution? A solution in any language will do.

编辑：有一点我应该更清楚，该算法必须返回false，如果交点超出线段的长度。 I.E.这是不是一种相贯线： lines.gif

A point I should have made clearer, the algorithm must return false if the point of intersection is beyond the lengths of the line segments. I.E. this isn't an intersecition: lines.gif

推荐答案

大多数问题的答案已经在这里似乎遵循的总体思路是：

Most of the answers already here seem to follow the general idea that:

找到两条直线传球给定的点的交集。 确定的交点属于这两个线段。

但是，当路口不经常发生，更好的方法可能是扭转这些步骤：

But when intersection does not occur often, a better way probably is to reverse these steps:

EX preSS中的 Y = AX + B 的（行通过A，B）和 Y = CX + D 的（行传递的形式直线C，D） 在看C和D是在同一侧的 Y = AX + B 的 在看，如果A和B是在同一侧的 Y = CX + D 的 如果答案以上都否，然后出现是的交集。否则就没有交集。 找到的交集，如果存在的话。 express the straight lines in the form of y = ax + b (line passing A,B) and y = cx + d (line passing C,D) see if C and D are on the same side of y = ax+b see if A and B are on the same side of y = cx+d if the answer to the above are both no, then there is an intersection. otherwise there is no intersection. find the intersection if there is one.

请注意：做第2步，只是检查（赛扬 - 一（CX） - b）和（240 - 一（DX） - B）具有相同的符号。步骤3是类似的。步骤5是从两个等式只是标准的数学。

Note: to do step 2, just check if (C.y - a(C.x) - b) and (D.y - a(D.x) - b) have the same sign. Step 3 is similar. Step 5 is just standard math from the two equations.

此外，如果你需要在每个线段与（N-1）等线段，precomputing步骤1中的所有线路节省您的时间进行比较。

Furthermore, if you need to compare each line segment with (n-1) other line segments, precomputing step 1 for all lines saves you time.