如何连接两个平行的2d多边形以创建无缝的3d网格? [英] How can I connect two parallel 2d polygons to create a seamless 3d mesh?
问题描述
假设我有两个多边形，一个像这样在另一个之上:
Suppose I have two polygons, one just above the other like this:
我想将它们的顶点连接起来，以围绕其周边的三角形创建一个3d网格.这张图片显示了您可以执行此操作的一种方法(橙色线代表三角形的边缘):
I'd like to connect up their vertices to create a 3d mesh out of triangles around their perimeters. This picture shows one way you might do this (orange lines represent triangle edges):
这种事情可以由人类直观地完成，但是将其转换为有效的算法确实很麻烦.
This sort of thing can be done intuitively by a human, but I'm having real trouble translating it into a working algorithm.
多边形存储为List<Vector2>.它们将始终是简单的，并且可能是凹入的.
The polygons are stored as a List<Vector2>. They will always be simple, and may be concave.
推荐答案
我会这样:
-
在每个多边形中找到两个最接近的点
这将用作起点
重新排列顶点
具有相同的缠绕规则，例如图像上的 CW
with the same winding rule for example CW like on image
为每个多边形找到中心"点
例如，所有顶点的平均值或边界框的中点或其他值.它不需要很精确，但是在复杂的凹形上，错误选择的中心位置会导致输出网格中的错误.
for example average of all vertexes or mid point of bounding box or whatever. It does not need to be precise but on complex concave shapes wrongly selected center position will cause errors in output mesh.
为每个顶点添加角度信息
只需使用
a=atan2(vertex-center)
毕竟，它应该像这样:
对于角度[deg]
index: 0 1 2 3 4
green: 355 085 135 170 230
blue: 020 135 255
现在，您只需将一个多边形中的2个最接近的顶点匹配到另一个多边形
Now you just match 2 closest vertexes from one polygon to the other
请不要忘记角度差最大为+/-180 deg，而且如果使用弧度，则将常量180,360转换为PI,2.0*PI
do not forget that angle difference is max +/-180 deg and also if you use radians then convert constants 180,360 to PI,2.0*PI
da=ang1-ang0;
while (da<-180.0) da+=360.0;
while (da>+180.0) da-=360.0;
if (da<0.0) da=-da;
对于下一个文本line(i,j)表示绿色的i-th顶点和蓝色多边形的j-th顶点
for next text line(i,j) will mean i-th vertex from green and j-th vertex from blue polygon
现在加入:
-
加入最近的顶点
只需处理所有绿色顶点并将它们连接到最接近的蓝色顶点(图像上的黑线)
just process all green vertexes and join them to closest blue vertex (black lines on image)
line(0,0)
line(1,1)
line(2,1)
line(3,1)
line(4,2)
填充孔
网格三角剖分每个顶点至少需要2个关节，因此处理点的连接较少:
mesh triangulation need at least 2 joints per vertex so process points that have less connections:
index: 0 1 2 3 4
green connections: 1 1 1 1 1
blue connections: 1 3 1
因此现在处理少于2次的蓝色顶点(0,2)，并将它们连接到最接近的绿色顶点(图像上的黄线)，而忽略已使用的连接(在这种情况下，选择下一个)
so now process less then 2 times used blue vertexes (0,2) and join them to closest green vertex (yellow lines on image) ignoring already used connection (chose next one in that case)
line(1,0)
line(3,2)
如此:
index: 0 1 2 3 4
green connections: 1 2 1 2 1
blue connections: 2 3 2
现在处理少于2倍使用过的顶点的其余绿色，再加入少于3倍使用过的蓝色顶点的绿色.如果连接数少于3，则选择已使用的连接的下一个点，如果连接数少于1，则选择最接近的点(图像上的红线).
now process the rest of green less then 2 times used vertexes joined to less then 3 times used blue vertex. Chose just the next point to already used connection if it has less then 3 connections and if there are more then 1 option use the closest one (red lines on image).
line(0,2)绿色0连接到蓝色0，因此从蓝色1,2中进行选择(太用了1，所以2)
line(2,-)绿色2被连接到蓝色1，蓝色1被使用了3次，因此忽略此顶点
line(4,-)绿色4被连接到蓝色2，蓝色2被使用了3次，因此请忽略此顶点
line(0,2) green 0 is connected to blue 0 so choose from blue 1,2 (1 is too used so 2)
line(2,-) green 2 is connected to blue 1 which is 3 times used so ignore this vertex
line(4,-) green 4 is connected to blue 2 which is 3 times used so ignore this vertex
index: 0 1 2 3 4
green connections: 2 2 1 2 1
blue connections: 2 3 3
仅此而已.现在，您应该像这样完成镶嵌处理:
and that is it all. Now you should have the tesselation done like this:
[notes]
- 您还可以使用周长位置/周长代替角度(有时效果更好)
- 凹面多边形可能使问题更加混乱，因此应添加一些三角形相交检查以免出现问题.
- 在多边形之间，顶点的数量也不应相差太多，在这种情况下，您仍然可以划分一些线以添加缺失点，否则3倍的使用条件将是错误的(如果顶点数相差2倍或2倍以上)
- 为了安全起见，应将多边形切成凸形部分，然后分别对其进行镶嵌处理，然后再将网格重新结合在一起
[Edit1]较不易受凹度和点密度影响的新方法
它的速度较慢，但看起来可以用于更复杂的形状.例如，我从评论中选择了修饰的星号和加号.
Its slower but looks its working for more complex shapes. As example I took the modified star and plus sign from the comments.
-
在每个多边形中找到两个最接近的点
这将用作起点
重新排列顶点
具有相同的缠绕规则，例如图像上的CW
with the same winding rule for example CW like on image
准备边缘列表
我们将需要一个结构来保存多边形之间的所有连接.我决定要这样:
we will need a structure holding all the connections between polygons. I decided to something like this:
List< List<int> > edg0,edg1;
其中edg0[point0][i]保持连接到point0的i-th point1.当心代码数组索引是点索引，数组的实际值n是全局点表pnt中的点索引，在这里我可以像这样在两者之间进行转换:
where edg0[point0][i] is holding i-th point1 connected to point0. Beware in the code array index is the point index and the actual value n the array is point index in a global point table pnt where I can transform between the two like this:
point0 = (pnt_index - ix0)/3
point1 = (pnt_index - ix1)/3
pnt_index = ix0 + 3*point0
pnt_index = ix1 + 3*point1
其中ix0,ix1是pnt中每个多边形的起始索引.
where ix0,ix1 are start indexes in pnt for each polygon.
加入封闭点
我没有加入最接近的点，而是添加了阈值，因此点距离必须小于阈值minll.此阈值是2个起点的距离(最小距离)的倍数.在代码中:
Instead of joining to the closest points I added thresholding so the point distance must be smaller than threshold minll. This threshold is a multiply of the distance of the 2 start points (min distance). In code:
minll=2.5*l0;
但是请注意，我正在比较distance^2的速度，所以如果您有距离就可以了
but beware I am comparing distance^2 for speed so if you got distance it would be
minl=sqrt(2.5)*l0;
可以调整乘法系数...
The multiply coefficient can be tweaked ...
如您所见，由于阈值作用，恒星上的遥远点未连接.
as you can see the distant point on star is not connected due thresholding.
在点0中填充未使用的孔
只需找到未使用的点0的序列，然后使用连接的第一个邻居从开始到结束对连接进行插值.因此，如果点0 i0 .. i1未连接并且它们的最近邻居已连接，则取其最近的连接点1 j0 .. j1并简单地将点i与j线性连接，其中:
simply find sequence of unused points0 and then interpolate the join from start to end with first neighbors that are connected. so if points0 i0 .. i1 are unconnected and their closest neighbors are connected take their closest connected points1 j0 .. j1 and simply linearly connect point i with j where:
i = i0 + t*(i1-i0)
j = j0 + t*(j1-j0)
其中，t是参数t=<0.0,1.0>，它通过所有整数"点索引. (使用DDA).
where t is parameter t=<0.0,1.0> going through all "integer" point indexes. (use DDA).
在点1中填充未使用的孔
与#5 相同，但在另一个多边形中查找未连接的点.
its the same as #5 but look for the unconnected points1 in the other polygon.
查找并连接单个连接线
两个端点仅使用一次.因此，只需找到这种边缘并将其连接到相邻点即可.
both endpoints are used just once. So simply find such edge and connect it to neighboring points.
查找形成QUAD孔的单数poinst0并对其进行三角测量
find singular poinst0 that form QUAD hole and triangulate it
，因此找到仅连接一次的point0，然后测试其邻居是否已连接回第一个已连接的point1.如果找不到，请对三角孔进行三角剖分.
so find point0 that is connected just once and then test its neighbors if the are connected back to the point1 the first one was connected to. If no you found QUAD hole so triangulate it.
现在只需从edg0,edg1
now just extract triangles and lines from the edg0,edg1
线很简单，因为它们已经直接编码，对于三角形，只需搜索相邻的point0以连接到同一point1.如果发现形成三角形.三角形也应在其他边列表中形成，因此也要搜索连接到同一point0的相邻point1.
line are simple as they are already encoded directly, for triangles simply search neighboring point0 for connection to the same point1. If found form a triangle. the triangles should be formed also in the other edge list so search neighboring point1 that are connected to the same point0 too.
以下是GL/C ++示例
List<double> pnt;
List<int> lin,tri;
int iy0=0,iy1=0,iy2=0,iy3=0;// pol0<iy0,iy1),pol1<iy1,iy2),merge<iy2,iy3)
int ix0=0,ix1=0,ix2=0; // points1<ix0,ix1), points2<ix1,ix2)
//---------------------------------------------------------------------------
void obj_init()
{
// input data
const double d=0.5; // distance between polygons
const double pol0[]={0.0,2.0, 1.0,2.0, 1.0,3.0, 2.00,3.0, 2.0,2.0, 3.00,2.0, 3.0,1.0, 2.0,1.0, 2.0,0.0, 1.0,0.0, 1.0,1.0, 0.0,1.0, 0.0,2.0};
// const double pol1[]={0.0,0.0, 1.0,1.0, 0.0,2.0, 1.25,1.5, 1.5,2.5, 1.75,1.5, 3.0,2.0, 2.0,1.0, 3.0,0.0, 1.5,0.7};
const double pol1[]={0.0,0.0, 1.0,1.0, 0.0,2.0, 1.25,1.5, 1.5,5.0, 1.75,1.5, 3.0,2.0, 2.0,1.0, 3.0,0.0, 1.5,0.7};
// ***
// variables
List<double> tmp;
List< List<int> > edg0,edg1; // edge table
double minll; // max distance to points that will join automatically
double p[3],l0,l;
int i,i0,i1,ii,ii0,ii1,di;
int j,j0,j1,jj,jj0,jj1,dj;
int k,n0,n1,e,de;
pnt.num=0;
lin.num=0;
// copy pol0 to point list pnt[]
ix0=pnt.num;
n0=sizeof(pol0)/sizeof(pol0[0]);
for (j=pnt.num,i=0;i<n0;)
{
pnt.add(pol0[i]); i++;
pnt.add(pol0[i]); i++;
pnt.add(-d);
} ix1=pnt.num; n0/=2;
// copy pol1 to point list pnt[]
n1=sizeof(pol1)/sizeof(pol1[0]);
for (j=pnt.num,i=0;i<n1;)
{
pnt.add(pol1[i]); i++;
pnt.add(pol1[i]); i++;
pnt.add(+d);
} ix2=pnt.num; n1/=2;
// add polygon edges
iy0=lin.num;
for (j=ix1-3,i=ix0;i<ix1;j=i,i+=3){ lin.add(j); lin.add(i); } iy1=lin.num;
for (j=ix2-3,i=ix1;i<ix2;j=i,i+=3){ lin.add(j); lin.add(i); } iy2=lin.num;
// find closest points -> start points i0,j0
i0=-1; j0=-1; l0=0.0; minll=0.0;
for (i=ix0;i<ix1;i+=3)
for (j=ix1;j<ix2;j+=3)
{
vector_sub(p,pnt.dat+i,pnt.dat+j);
l=vector_len2(p);
if ((i0<0)||(l<l0)){ l0=l; i0=i; j0=j; }
} minll=2.5*l0;
// reorder points so closest ones are first
if (i0!=ix0)
{
tmp.num=0; for (i=ix0;i<ix1;i++) tmp.add(pnt.dat[i]); // copy to temp
for (j=i0,i=ix0;i<ix1;i++,j++,(j>=ix1)?j=ix0:j=j) pnt.dat[i]=tmp.dat[j-ix0]; // reorder
}
if (j0!=ix1)
{
tmp.num=0; for (i=ix1;i<ix2;i++) tmp.add(pnt.dat[i]); // copy to temp
for (j=j0,i=ix1;i<ix2;i++,j++,(j>=ix2)?j=ix1:j=j) pnt.dat[i]=tmp.dat[j-ix1]; // reorder
}
// init edge lists
edg0.allocate(n0); edg0.num=n0; for (i=0;i<n0;i++) edg0[i].num=0;
edg1.allocate(n1); edg1.num=n1; for (i=0;i<n1;i++) edg1[i].num=0;
// join close points
for (ii=0,i=ix0;i<ix1;i+=3,ii++)
{
j0=-1; jj0=-1; l0=0.0;
for (jj=0,j=ix1;j<ix2;j+=3,jj++)
{
vector_sub(p,pnt.dat+i,pnt.dat+j);
l=vector_len2(p);
if ((j0<0)||(l<l0)){ l0=l; j0=j; jj0=jj; }
}
if ((j0>=0)&&(l0<minll))
{
edg0.dat[ii ].add(j0);
edg1.dat[jj0].add(i);
}
}
// fill unused holes in points0
for (e=1,i=0;e;)
{
e=0;
// find last used point0 -> i0
i0=-1; i1=-1;
for (j=0;j<n0;j++,i++,(i==n0)?i=0:i=i) if (edg0.dat[i].num) i0=i; else if (i0>=0) break;
// find next used point0 -> i1
if (i0>=0) if (edg0.dat[i].num==0)
for (j=0;j<n0;j++,i++,(i==n0)?i=0:i=i) if (edg0.dat[i].num){ i1=i; break; }
if (i1>=0)
{
// find last used point1 joined to i0 -> j0
j0=-1; jj0=-1; l0=0.0;
i=i0+1; if (i>=n0) i=0; ii=ix0+i+i+i;
for (j=0;j<edg0.dat[i0].num;j++)
{
jj=edg0.dat[i0].dat[j];
vector_sub(p,pnt.dat+ii,pnt.dat+jj);
l=vector_len2(p);
if ((j0<0)||(l<l0)){ l0=l; j0=(jj-ix1)/3; jj0=jj; }
} i0=i;
// find next used point1 joined to i1 -> j1
j1=-1; jj1=-1; l0=0.0;
i=i1-1; if (i<0) i=n0-1; ii=ix0+i+i+i;
for (j=0;j<edg0.dat[i1].num;j++)
{
jj=edg0.dat[i1].dat[j];
vector_sub(p,pnt.dat+ii,pnt.dat+jj);
l=vector_len2(p);
if ((j1<0)||(l<l0)){ l0=l; j1=(jj-ix1)/3; jj1=jj; }
} i1=i;
// join i0..i1 <-> j0..j1
di=i1-i0; if (di<0) di+=n0; // point0 to join
dj=j1-j0; if (dj<0) dj+=n1; // point1 to join
de=di; if (de<dj) de=dj; // max(points0,point1)
for (e=0;e<=de;e++)
{
i=i0+((e*di)/de); if (i>=n0) i-=n0; ii=ix0+i+i+i;
j=j0+((e*di)/de); if (j>=n1) j-=n1; jj=ix1+j+j+j;
for (k=0;k<edg0.dat[i].num;k++) // avoid duplicate edges
if (edg0.dat[i].dat[k]==jj)
{ k=-1; break; }
if (k>=0)
{
edg0.dat[i].add(jj);
edg1.dat[j].add(ii);
}
}
e=1;
}
}
// fill unused holes in points1
for (e=1,i=0;e;)
{
e=0;
// find last used point1 -> i0
i0=-1; i1=-1;
for (j=0;j<n1;j++,i++,(i==n1)?i=0:i=i) if (edg1.dat[i].num) i0=i; else if (i0>=0) break;
// find next used point1 -> i1
if (i0>=0) if (edg1.dat[i].num==0)
for (j=0;j<n1;j++,i++,(i==n1)?i=0:i=i) if (edg1.dat[i].num){ i1=i; break; }
if (i1>=0)
{
// find last used point0 joined to i0 -> j0
j0=-1; jj0=-1; l0=0.0;
i=i0+1; if (i>=n1) i=0; ii=ix1+i+i+i;
for (j=0;j<edg1.dat[i0].num;j++)
{
jj=edg1.dat[i0].dat[j];
vector_sub(p,pnt.dat+ii,pnt.dat+jj);
l=vector_len2(p);
if ((j0<0)||(l<l0)){ l0=l; j0=(jj-ix0)/3; jj0=jj; }
} i0=i;
// find next used point0 joined to i1 -> j1
j1=-1; jj1=-1; l0=0.0;
i=i1-1; if (i<0) i=n1-1; ii=ix1+i+i+i;
for (j=0;j<edg1.dat[i1].num;j++)
{
jj=edg1.dat[i1].dat[j];
vector_sub(p,pnt.dat+ii,pnt.dat+jj);
l=vector_len2(p);
if ((j1<0)||(l<l0)){ l0=l; j1=(jj-ix0)/3; jj1=jj; }
} i1=i;
// join i0..i1 <-> j0..j1
di=i1-i0; if (di<0) di+=n1; // point1 to join
dj=j1-j0; if (dj<0) dj+=n0; // point0 to join
de=di; if (de<dj) de=dj; // max(points0,point1)
for (e=0;e<=de;e++)
{
i=i0+((e*di)/de); if (i>=n1) i-=n1; ii=ix1+i+i+i;
j=j0+((e*di)/de); if (j>=n0) j-=n0; jj=ix0+j+j+j;
for (k=0;k<edg1.dat[i].num;k++) // avoid duplicate edges
if (edg1.dat[i].dat[k]==jj)
{ k=-1; break; }
if (k>=0)
{
edg1.dat[i].add(jj);
edg0.dat[j].add(ii);
}
}
e=1;
}
}
// find&connect singular connection lines (both endpoints are used just once)
for (i=0;i<n0;i++)
if (edg0.dat[i].num==1) // point0 used once
{
jj=edg0.dat[i].dat[0]; // connected to
j=(jj-ix1)/3;
if (edg1.dat[j].num==1) // point1 used once
{
i0=i-1; if (i< 0) i+=n0; // point0 neighbors
i1=i+1; if (i>=n0) i-=n0;
ii =ix0+i +i +i;
ii0=ix0+i0+i0+i0;
ii1=ix0+i1+i1+i1;
if (edg1.dat[j].dat[0]!=ii0) // avoid duplicate edges
{
edg1.dat[j].add(ii0);
edg0.dat[i0].add(jj);
}
if (edg1.dat[j].dat[0]!=ii1) // avoid duplicate edges
{
edg1.dat[j].add(ii1);
edg0.dat[i1].add(jj);
}
}
}
// find singular poinst0 that form QUAD hole and triangulate it
for (i=0;i<n0;i++)
if (edg0.dat[i].num==1) // point0 used once
{
jj=edg0.dat[i].dat[0]; // connected to
j=(jj-ix1)/3;
i0=i-1; if (i< 0) i+=n0; // point0 neighbors
i1=i+1; if (i>=n0) i-=n0;
j0=j-1; if (j< 0) j+=n1; // point1 neighbors
j1=j+1; if (j>=n1) j-=n1;
ii =ix0+i +i +i;
ii0=ix0+i0+i0+i0;
ii1=ix0+i1+i1+i1;
jj0=ix1+j0+j0+j0;
jj1=ix1+j1+j1+j1;
for (k=0;k<edg1.dat[j].num;k++) // avoid duplicate edges
{
if (edg1.dat[j].dat[k]==ii0) i0=-1;
if (edg1.dat[j].dat[k]==ii1) i1=-1;
}
if (i0>=0)
{
edg1.dat[j ].add(ii0);
edg0.dat[i0].add(jj );
}
if (i1>=0)
{
edg1.dat[j ].add(ii1);
edg0.dat[i1].add(jj );
}
}
// extract merge triangles from edge0
for (i0=0,i1=1;i0<n0;i0++,i1++,(i1>=n0)?i1=0:i1=i1)
{
ii0=ix0+i0+i0+i0;
ii1=ix0+i1+i1+i1;
// find point1 joined to both points0 i0,i1
for (i=0;i<edg0.dat[i0].num;i++)
for (j=0;j<edg0.dat[i1].num;j++)
if (edg0.dat[i0].dat[i]==edg0.dat[i1].dat[j])
{
jj=edg0.dat[i0].dat[i];
tri.add(ii0);
tri.add(ii1);
tri.add(jj);
}
}
// extract merge triangles from edge1
for (i0=0,i1=1;i0<n1;i0++,i1++,(i1>=n1)?i1=0:i1=i1)
{
ii0=ix1+i0+i0+i0;
ii1=ix1+i1+i1+i1;
// find point1 joined to both points0 i0,i1
for (i=0;i<edg1.dat[i0].num;i++)
for (j=0;j<edg1.dat[i1].num;j++)
if (edg1.dat[i0].dat[i]==edg1.dat[i1].dat[j])
{
jj=edg1.dat[i0].dat[i];
tri.add(ii0);
tri.add(ii1);
tri.add(jj);
}
}
// extract merge edges
for (i=ix0,ii=0;ii<n0;ii++,i+=3)
for (j=0;j<edg0.dat[ii].num;j++)
{
lin.add(i);
lin.add(edg0.dat[ii].dat[j]);
}
iy3=lin.num;
/*
// [debug]
AnsiString txt="";
for (txt+="[edg0]\r\n",i=0;i<n0;i++,txt+="\r\n")
for (txt+=AnsiString().sprintf("%2i:",i),j=0;j<edg0.dat[i].num;j++)
txt+=AnsiString().sprintf("%2i ",(edg0.dat[i].dat[j]-ix1)/3);
for (txt+="\r\n[edg1]\r\n",i=0;i<n1;i++,txt+="\r\n")
for (txt+=AnsiString().sprintf("%2i:",i),j=0;j<edg1.dat[i].num;j++)
txt+=AnsiString().sprintf("%2i ",(edg1.dat[i].dat[j]-ix0)/3);
e=FileCreate("debug.txt");
FileWrite(e,txt.c_str(),txt.Length());
FileClose(e);
*/
}
//---------------------------------------------------------------------------
void obj_draw()
{
int i,j;
glLineWidth(2.0);
glColor3f(0.1,0.1,0.1); glBegin(GL_LINES); for (i=0;(i<iy0)&&(i<lin.num);i++) glVertex3dv(pnt.dat+lin.dat[i]); glEnd(); // aux
glColor3f(0.1,0.1,0.9); glBegin(GL_LINES); for ( ;(i<iy1)&&(i<lin.num);i++) glVertex3dv(pnt.dat+lin.dat[i]); glEnd(); // polygon 0
glColor3f(0.9,0.1,0.1); glBegin(GL_LINES); for ( ;(i<iy2)&&(i<lin.num);i++) glVertex3dv(pnt.dat+lin.dat[i]); glEnd(); // polygon 1
glColor3f(0.1,0.9,0.1); glBegin(GL_LINES); for ( ;(i<iy3)&&(i<lin.num);i++) glVertex3dv(pnt.dat+lin.dat[i]); glEnd(); // merge
glColor3f(0.1,0.1,0.1); glBegin(GL_LINES); for ( ; (i<lin.num);i++) glVertex3dv(pnt.dat+lin.dat[i]); glEnd(); // aux
glLineWidth(1.0);
glPointSize(5.0);
glColor3f(0.9,0.9,0.1); glBegin(GL_POINTS); glVertex3dv(pnt.dat+ix0+0); glVertex3dv(pnt.dat+ix1+0); glEnd(); // 1st points
glColor3f(0.1,0.9,0.9); glBegin(GL_POINTS); glVertex3dv(pnt.dat+ix0+3); glVertex3dv(pnt.dat+ix1+3); glEnd(); // 2nd points
glColor3f(0.3,0.3,0.3); glBegin(GL_POINTS); for (i=0;i<pnt.num;i+=3) glVertex3dv(pnt.dat+i); glEnd(); // points
glPointSize(1.0);
glDisable(GL_CULL_FACE);
glColor3f(0.2,0.2,0.2); glBegin(GL_TRIANGLES); for (i=0;i<tri.num;i++) glVertex3dv(pnt.dat+tri.dat[i]); glEnd(); // faces
}
//---------------------------------------------------------------------------
我还使用了我的动态列表模板，所以:
I also use mine dynamic list template so:
List<double> xxx;与double xxx[];相同
xxx.add(5);将5添加到列表的末尾
xxx[7]访问数组元素(安全)
xxx.dat[7]访问数组元素(不安全但快速的直接访问)
xxx.num是数组的实际使用大小
xxx.reset()清除数组并设置xxx.num=0
xxx.allocate(100)为100个项目预分配空间
List<double> xxx; is the same as double xxx[];
xxx.add(5); adds 5 to end of the list
xxx[7] access array element (safe)
xxx.dat[7] access array element (unsafe but fast direct access)
xxx.num is the actual used size of the array
xxx.reset() clears the array and set xxx.num=0
xxx.allocate(100) preallocate space for 100 items
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