usaco 计算几何

usaco 计算几何

usaco计算几何2010-05-10 09：09Computational Geometry Prerequisites Graph TheoryShortest PathTools This module discusses several algorithms that calculate various geometric properties,mostly based on only two operations described below：cross product and arctangent.

Cross Product The cross product of uand vis written as ux

https://www.360docs.net/doc/346593229.html,putationally,the cross product of two three-dimensional vectors uand vis the vector determinant of the following matrix(where i,j,and kare unit vectors in the x,y,and zdirections respectively)：|i jk||ux uy uz||vx vy vz|That equation works out to：

(uyvz-vyuz)i+(uzvx-uxvz)j+(uxvy-uyvx)k This definition can be used for vectors in two dimensions by using three-dimensional vectors with az component of 0.The resulting vector will only have az value.

The cross product has three properties：

The cross product of two vectors is perpendicular to both vectors.The length of the cross product is equal to the product of：

the length of u,the length of v,andthe sine of the angle between the vectors.Of the two different directions that are perpendicular to both uand v,the direction the cross product points depends on whether uis``to the right''of vor``to the left.''

Dot product The dot product of two vectors uand vis ascalar written as u·https://www.360docs.net/doc/346593229.html,putationally,it is defined in three dimensions as：uxvx+u yvy+uzv z

The dot product is actually equal to the product of：

the length of uthe length of vthe cosine of the angle between uand v.Presuming uand vare non-zero,if the dot product if negative,u and vmake an angle greater than 90 degrees.If it is zero,then uand vare perpendicular.If ucdot vis positive,then the two vectors form an acute angle.

Arctangent The arctangentfunction calculates the(an)angle whose tangent is its argument and generally returns areal number between-

pi/2 and pi/2.An additional function in C,atan2,takes two arguments：a DELTA yvalue and aDELTA xvalue(in that order！).It determines the angle between the given vector and the positive xaxis and returns avalue between-pi and pi.This has the advantage of removing concerns about dividing by zero or writing code to repair angles in order to handle the negative xcases.The atan2 function is almost always easier to use than the simpler atan function that takes only one argument.

Particular Debugging Problems The main problem with geometric problems is that they spawn alot of special cases.Be on the lookout for these special cases and make sure your program works for all of them.

Floating point calculations also create anew set of

problems.Floating point calculations are rarely precise,as the computer only maintains so many bits(digits)of accuracy：be aware of this.In particular,when checking if two values are equal,check to see if they are within some small tolerance of each other not precisely equal.

Geometric Algorithms Here are some of snippets that can help you solve geometry problems.

Area of Triangle To cal culate the area of atriangle with

vertices(a,b,c),pick avertex(say a)and create avector to the other two vertices(let u=b-a,and v=c-a).The area of the triangle(a,b,c)is

one half the length of cross product ux v.An alternative method to find the area of triangle is to use Hero's formula.If the lengths of the sides of atriangle are a,b,and c,let s=(a+b+c)/2.The area of the triangle is then sqrt(s*(s-a)*(s-b)*(s-c)).Are Two Line Segments Parallel?

To check if two line segments are parallel,create vectors along each line segment and check to see if their cross product

is(almost)zero.

Area of polygon The area of apolygon with vertices(x 1,y 1),.,(x n,y n)is equal to the determinant：1|x1 x2.xn|---||2|y1 y2.yn|

where the determinate is defined to be similar to the 2by

2determinant：x1 y2+x2y3+.+xn y1-y1 x2-y2x3-.-yn x1 Distance from apoint to aline The distance from apoint Pto aline AB is given by the magnitude of the cross product.In particular,d(P,AB)=|(P-A)x(B-

A)|/|B-A|.

To determine the distance from apoint Pto the plane defined by A,B,and C,let n=(B-A)x(C-A).The distance is then give by the

following equation：d(P,ABC)=(P-A)·n/|n|.

Points on aline Apoint is on aline if the distance from the point to the line is 0.

Points on the same side of line This notion only makes sense for two dimensions.To check if points Cand Dare on the same side of line AB,calculate the zcomponent of(B-A)x(C-A)and(B-A)x(D-A).If the zcomponents have the same sign(i.e.,their product is positive),then Cand Dare on the same side of the line AB.

Point on line segment To calculate if apoint Cis on the line segment AB,check if Cis on the line AB.If it is,then check if the length of AB is equal to the sum of the lengths of AC and CB.

Point in triangle To check if apoint Ais in atriangle,find another point Bwhich is within the triangle(the average of the three vertices works well).Then,check if the point Ais on the same side of the three lines defined by the edges of the triangle as B.

Point in convex polygon The same trick works for aconvex polygon：

Four(or more)points are coplanar To determine if acollection of points is coplanar,select three points,A,B,and C.Now,if,for any other point D,(B-A)x(C-A))·(D-A)=~0,then the collection of points resides in some plane.

Two lines intersect Two lines intersect if and only if they are not parallel in two dimensions.

In three dimensions,two lines AB and CD intersect if they are not parallel and A,B,C,and Dare coplanar.

Two line segments intersect In two dimensions,two line segments AB and CD intersect if and only if Aand Bare on opposite sides of the line CD and Cand Dare on opposite sides of line AB.Note that both of the checks are necessary,as for the last case one of the checks returns true,while the other testifies to the fact that AB and CD do not intersect.In three dimensions,solve following system of equations,where iand jare the unknowns：

Ax+(Bx-Ax)i=Cx+(Dx-Cx)j Ay+(By-Ay)i=Cy+(Dy-Cy)j Az+(Bz-

Az)i=Cz+(Dz-Cz)j If this system has asolution(i,j),where 0=i=1 and

0=j=1,then the line segments intersect at：(Ax+(Bx-Ax)i,Ay+(By-

Ay)i,Az+(Bz-Az)i.

Point of Intersection of Two Lines For the lines AB and CD in two dimensions,the most straight-forward way to calculate the

intersection of them is to solve the system of two equations and two unknowns：

Ax+(Bx-Ax)i=Cx+(Dx-Cx)j Ay+(By-Ay)i=Cy+(Dy-Cy)jThe point of intersection is：

(Ax+(Bx-Ax)i,Ay+(By-Ay)i)In three dimensions,solve the same system of equations as was used to check line intersection,and the point of intersection is：

(Ax+(Bx-Ax)i,Ay+(By-Ay)i,Az+(Bz-Az)i)Checking convexity of 2-dimensional polygon To check the convexity of a2-dimensional polygon,walk the polygon in clock-wise order.For each triplet of consecutive points(A,B,C),calculate the cross product(B-A)x(C-A).If the zcomponent of each of these vectors is positive,the polygon is convex.

Point in non-convex polygon To calculate if apoint is within anonconvex polygon,make aray from that point in arandom direction and count the number of times it intersects the polygon.If the ray intersects the polygon at avertex or along an edge,pick anew direction.Otherwise,the point is within the polygon if and only if th eray intersects the polygon an odd number of times.

This method also extends to three dimensions(and higher),but the restriction on intersection is that it only intersects at faces and not at either avertex or an edge.

Geometry Methodologies Geometric problems introduce several different tricks that can be used to either reduce the run-time or approximate the solution.

Monte Carlo The first geometric trick is based on

randomness.Instead of calculating the probability that something occurs,simulate arandom event and calculate the fraction of times it occurs.If enough events are simulated,the difference between these two values becomes very small.

This can be helpful to determine something like the area of afigure.Instead of calculating the area directly,determine abounding box,and throw``darts''at the box,and estimate what the probability of hitting the figure is.If this is calculated accurately enough,this can give agood estimate of the actual area.

The problem with this method is to get agood relative error(error divided by the actual value)requires alarge number of successful events.If the probability of the event occurring is very small,the method does not yield good results.

Partitioning Partitioning is amethod to improve the speed of ageometric algorithm.This entails dividing the plane up into

sections(usually by agrid but sometimes into radial sections or some other method),and bucketing the objects into appropriate

section(s).When looking for objects within some figure,only those sections which have anon-zero intersection with that figure need to be examined,thereby greatly reducing the cost of the algorithm.This is helpful to determine the set of objects within some distance of agiven point(the figure is acircle)or to check for intersections(the figure is aline).

Graph Problems Sometimes what may look like ageometric problem is really agraph problem.Just because the input is points in the plane does not mean it's ageometric algorithm.

Example Problems Poi nt Moving Given aset of line segments in the plane,and two points Aand B,is it possible to move from Ato Bwithout crossing any of the segments?

The line segments partition the plane into regions.Determine these regions,and see if Aand Breside in the same region.

Bicycle Routing Given acollection of non-intersecting buildings along with start and end locations,find the shortest path from Ato Bthat doesn't go through any buildings.

Analysis：This is really agraph problem.The nodes are the start and end locations,along with the vertices of the buildings.There are edges between any two nodes such that the line segment between them does not intersect any buildings,with weight equal to the length of the length of the line segments.Once that graph has been

calculated,the problem is shortest path.

Maximizing Line Intersections Given acollection of segments in the plane,find the greatest number of segments which can by intersected by drawing asingle line.

Analysis：With alittle bit of thought,it is clear that the line segment must pass through two of the vertices of the collection of line segments.Thus,try all pairs of vertices,and cal culate the crossing for https://www.360docs.net/doc/346593229.html,bining this with partitioning gives an algorithm that runs fairly quickly.

Polygon Classification Given acollection of segments defining apolygon,determine if it is simple(no two non-consecutive line segments intersect)and convex.

计算机应用中的解析几何

译by zhougu

学习该内容的基础

*图论问题

*最短路径问题

工具

这个模块论述了一些具有几何特性的多方面做计划的算法，而这些算法大多都建立在以下的两个概念的基础上：向量乘积和反正切。

向量的乘积：

向量u和v的乘积被记做u xv。而对于计算机程序来说，两个三维空间内的向量u和的积是决定于下面的这个矩阵。(i，j，k是单位向量，而x，y，z 分别是它们的方向)

|i jk|

|ux uy uz|

|vx vy vz|可以通过这个方程计算出：(uyvz-vyu z)i+(uzvx-u

xvz)j+(uxvy-uvx)k

这个方法完全可以被用在二维的空间内，那时我们认为z=0，自然结果向量中的z也为0。

向量的乘法有三个性质：

*两个向量的积在空间范围内同时垂直于这两个向量。

*向量u和v的积的长度等于向量u的长度、向量v的长度和两向量夹角的正弦值的乘积。

*在两个可能出现的方向中，向量u和向量v的积的方向取决于u和v的位置关系。

向量的数量积：

向量u和向量v的数量积记做u?v，用与刚才类似的方程表示为：u xvx+u yvy+u zvz

而事实上我们可以用标量u的长度、v标量的长度和两向量夹角的余弦值的乘积来表示向量u和向量v的数量积。如果向量u和向量v的夹角大于90度，而u和v都是非0的，那么它们的数量积是负数。如果它们的数量积等于0，那么它们是互相垂直的。如果u?v的结果是正数，那么它们夹的是一个锐角。

反正切：反正切作为函数通常可以通过正切值计算出一个处在-Pi/2到

Pi/2之间的角的度数。C语言中额外的atan2函数带有两个引数：DELTA y的值和DELTA x的值。它会测定所给向量与x轴的夹角并返回一个处在-Pi/2到Pi/2之间的角度。它的好处就在于消去了涉及到0做被除数和为处理多种角度情况时所写的代码。大多时候，函数atan2要比只带有一个引数的函数atan简单。

特殊的调试问题：

解决几何问题时最主要是问题是它们往往有很多特殊情况，所以要确认你的程序可以应付所有的特殊情况。

浮点数据的计算总会产生一系列的问题。浮点数据的计算很少是100%精确的，因为电脑本身也只能维持一定位数的准确度。尤其是判断两个值是否相等时，电脑只是看它们的差是否在一个很小的容差值的范围内。

几何算法：

以下是几个可以帮助我们解决几何问题的资料。

三角形区域：

计算一个顶点为(a,b,c)的三角形所在的区域，要先选择一个顶点(比如顶点a)，并创建两个与另外三角形的三边有如下关系的向量：使向量u=b-a，向量v=c-a。三角形a、b、c的区域就是向量u和向量v的乘积的1.5倍。

还有一个求三角形区域可以选择的方法是用hero定理。如果三角形三边的长度分别为a、b、c，设s=(a+b+c)/2，这个三角形的区域就等于：sqrt(s*(s-a)*(s-b)*(s-c))两条线段是否平行：

检验两条线段是否平行，可以沿着两条线段的方向创建两个向量，然后看它们的乘积是否为0。

多边形区域：

一个顶点为(x 1,y 1),.,(x n,y n)的多边形的区域等于下面这个行列式。

1|x1 x2.xn|

---||

2|y1 y2.yn|

这个行列阵的算法类似于2 X2的行列阵：x1 y2+x2y3+.+xn y1-y1 x2-

y2x3-.-yn x1。

点到直线的距离：

值得注意的是点P到直线AB的距离往往是由向量乘积量得出的：

d(P,AB)=|(P-A)x(B-A)|/|B-A|。要决定点P到A、B、C三点所在的平面的距离，需先设n=(B-A)x(C-A)，然后用这个公式来计算：d(P,ABC)=(P-

A)?n/|n|。

在直线上的点：

在直线上的点到直线的距离为0。

在直线同侧的点：

这个想法只对二维的空间有意义。要检验点C和点D是否在直线AB同侧，只需计算(B-A)x(C-A)和(B-A)x(D-A)的值。如果它们同号，C和D就在直线AB 的同侧。

三角形内的点：

检验点A是否在三角形内部，我们可以先在三角形内部找一个点B(三个顶点的平均值即可)。然后检验A相对三边是否都在B的同侧。

凸多边形内的点：

这和三角形是同样的做法。

多个(四个以上)点共面：

要判断一系列点是否共面，我们可以先选择三个点A、B、C，它们必然是共面的，在任意挑选一点D，如果(B-A)x(C-A))?(D-A)=0，则这个点与点A、B、C在一个平面内。

两线段相交：

在二维空间内，两条线段AB和CD相交，如果A、B分别在线段CD的两侧而且C、D分别在线段AB的两侧。

把这两个核对全记录下来往往是不必要的，如果其中一个的核对证明AB和CD是不相交的就返回false并结束核对。在三维空间内，利用以下的等式(i,j 不是已知的)：

Ax+(Bx-Ax)i=Cx+(Dx-Cx)j Ay+(By-Ay)i=Cy+(Dy-Cy)j Az+(Bz-

Az)i=Cz+(Dz-Cz)j

如果已知了(i,j)，且i、j在0和1之间，那么两线段相交于(Ax+(Bx-Ax)i,Ay+(By-Ay)i,Az+(Bz-Az)i。

两条直线交点：

要计算二维空间内的两直线AB、CD的交点，最容易想到的方法就是构造两直线方程：

Ax+(Bx-Ax)i=Cx+(Dx-Cx)j Ay+(By-Ay)i=Cy+(Dy-Cy)j

交点就是：

(Ax+(Bx-Ax)i,Ay+(By-Ay)i)

在三维空间内，解决的方法和上面的大致相同，交点是：

(Ax+(Bx-Ax)i,Ay+(By-Ay)i,Az+(Bz-Az)i)

检验二维多边形的凹凸：

要检验二维多边形的凹凸，可以按顺时针方向遍历整个多边形。对于每一个连续的3个顶点(A，B，C)，计算(B-A)x(C-A)的乘积。如果所有结果都是正的，那么这个多边形就是凸多边形。

凹多边形内的点：

要计算一个点是否在凹多边形内部，可以从该点做一条不定方向的射线，然后看它与这个凹多边形的交点的数目。如果它经过凹多边形的一个顶点或者平行于一条边，就换个方向重做射线。如果射线与凹多边形的交点数为奇数，那么该点在凹多边形内。

这个判定方法在三维(或者多维)，只是对交点的限制由顶点和边变成了平面。

几何方法论：

几何问题会引进很多减小程序运行时间或近似解决问题的技巧。

蒙特卡洛模拟法：

第一个几何技巧是建立在随机基础之上的。就是依照概率随机模拟一个事件产生的值来替代计算事件可能产生的值。如果模拟的事件足够多的话，那么这两个值之间的差就会变得很小。

这对于确定一些有考虑范围的问题有帮助。因为它不必求出准确的范围，只要先确定一个大致的范围，然后估计可能出现的结果，再在已有范围里做调整。如果这个范围足够精确，你就可以偷着高兴了。这个方法的问题是会产生一个相对的误差，所以考虑的可能的事件数越多越好。如果你对可能的事件考虑得太少，用这个方法是不会有好结果的。

分割法：

分割法是一种改进几何算法速度的方法。这种方法把平面拆分成很多部分(常常按格划分，有时也辐射状地划分或者用其他方式)，适当地把对象存入各部分中。然后只需处理那些存有对象的部分。从而大大减小了算法的花费。这个方法对那些求点的距离(整体是圆)或者交叉点(整体是线)的问题有帮助。

图的问题：

有时图的问题往往会被误认为是几何问题。几何问题是不能仅仅通过输入的是平面上的点来确定的。

例题：

点的移动：

给你一些平面内的线段和两个点A、B，问是否可能从A点到达B点而且不穿越任何线段。换种说法就是线段把整个平面分割成多个封闭的区域，问A和B是否在同一个区域内。

骑车路线：在起点与终点之间，给你一些不交叉的建筑物，求在不穿越建筑物的前提下，起点到终点的最短距离。

算法：这确实是一个图的问题，节点是起点和终点。如果与建筑物不交叉，在每两个建筑物的节点之间都可以生成一条线段，距离的增加量就是线段的长度；如果与建筑物交叉，就向终点的方向转。这样走一遍，得出来的就是要求的最短距离。

最少交点：

给你一些平面内的线段，尽可能少的做直线，使所给的线段全部被做出的直线穿过，求最少需要做直线的条数。

算法：考虑一下，思路应该很清楚。直线一定经过所有线段的端点中的两个，这样就可以玫举所有线段的两个端点，再计算每种情况下交点的个数。如果能把此算法和分割法一起使用，会大大提高算法的性能。

多边形种类：

给你一些平面内的线段来定义一个多边形，判断它是简单的(没有两条不连续的线段有交点)并判断它是否是凸多边形。

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