小弟遇到难题了，百思不得其解啊

幻天桥

最近想做一条多段线顶点压缩，条件如下：首先将一条多段线的首尾相连，然后求出个点到该连线的距离，求出最大值的点，用该最大值和事先给定的阀值比较，若大于阀值，则保留该点，若小于阀值，则删除该点，再将该点和首尾点各自相连，设起点到该点的连线为L1，该点到结束点的连线为L2，然后用同样的方法依次求出在起点到该点之间的点到L1的距离，求出最大值的点，同理，求出到L2最大距离的点，继续和阀值比较，依次类推，求出满足条件的所有的点，最后将这些点连成一条新的多段线，这就是压缩过的多段线，部分代码如下：//
#include "StdAfx.h"
#include "StdArx.h"
#include"acedads.h"
#include"adscodes.h"
#include <stdlib.h>
#include <rxobject.h>
#include <rxregsvc.h>
#include <aced.h>
#include <dbents.h>
#include <adslib.h>
#include <geassign.h>
#include "math.h"
double pt[50][3],h[50];
double v=10;
void listPline();
void iterate(AcDbObjectId id);
int maxi(int,int);
void studyaaddd()
{
listPline();
}
void
listPline()
{
int rc;
ads_name en;
AcGePoint3d pt;
rc = acedEntSel("\nSelect a polyline: ", en,
asDblArray(pt)); if (rc != RTNORM) {
acutPrintf("\nError during object selection");
return;
} AcDbObjectId eId;
acdbGetObjectId(eId, en); AcDbObject *pObj;
acdbOpenObject(pObj, eId, AcDb::kForRead);
if (pObj->isKindOf(AcDb2dPolyline::desc())) {
pObj->close();
iterate(eId);
} else {
pObj->close();
acutPrintf("\nSelected entity is not an AcDb2dPolyline. \nMake sure the setvar PLINETYPE is set to 0 before createing a polyline");
}
}
void
iterate(AcDbObjectId plineId)
{
AcDb2dPolyline *pPline;
acdbOpenObject(pPline, plineId, AcDb::kForRead); AcDbObjectIterator *pVertIter= pPline->vertexIterator();
pPline->close(); AcDb2dVertex *pVertex;
AcGePoint3d location;
AcDbObjectId vertexObjId;
int i=0;
for (int vertexNumber = 0;!pVertIter->done();
vertexNumber++,pVertIter->step())
{
vertexObjId = pVertIter->objectId();
acdbOpenObject(pVertex, vertexObjId,
AcDb::kForRead);
location = pVertex->position();
pt[0]=float(location[X]);
pt[1]=float(location[Y]);
pt[2]=float(location[Z]);
pVertex->close();

acutPrintf("\npt[%d][0]=%.4f,pt[%d][1]=%.4f,pt[%d][2]=%.4f",
i,location[X],i,location[Y],i,location[Z]);
i++;
} delete pVertIter;
acutPrintf("\n总共%d个顶点",i);
int m,n;
n=i-1;
double max1,a1,b1,c1,e1,s1;
b1=sqrt((pt[1][1]-pt[0][1])*(pt[1][1]-pt[0][1])+(pt[1][0]-pt[0][0])*(pt[1][0]-pt[0][0]));
a1=sqrt((pt[2][1]-pt[0][1])*(pt[2][1]-pt[0][1])+(pt[2][0]-pt[0][0])*(pt[2][0]-pt[0][0]));
c1=sqrt((pt[2][1]-pt[1][1])*(pt[2][1]-pt[1][1])+(pt[2][0]-pt[1][0])*(pt[2][0]-pt[1][0]));
e1=(a1+b1+c1)/2;
s1=sqrt(e1*(e1-a1)*(e1-b1)*(e1-c1));
max1=(2*s1)/a1;

switch(i)
{case 2:acutPrintf("\n不符合条件,无法求距离!");return;
case 3:acutPrintf("\n第2点距离为%f:",max1);
acutPrintf("\n无最大值!");
}
if(i>3)
{
acutPrintf("\n该点为pt[%d]且最大值h[%d]=%lf",m,m,h[m]);
for(m=maxi(0,n);p=maxi(0,n);n-p>1;m>1;m=maxi(0,m);p=maxi(p,n))
{k1=maxi(maxi(0,m),m);
u1=maxi(p,maxi(p,n));
for(k2=k1;k3=k1;m-k3>1;k2-maxi(0,m)>1;k2=maxi(maxi(0,m),k2);k3=maxi(k3,m))
}
}
int maxi(int x,int y)
{int z;
double a,b,c,e,s,max=0;
for(int j=x+1;j<y;j++)
{a=sqrt((pt[y][1]-pt[x][1])*(pt[y][1]-pt[x][1])+(pt[y][0]-pt[x][0])*(pt[y][0]-pt[x][0]));
b=sqrt((pt[j][1]-pt[x][1])*(pt[j][1]-pt[x][1])+(pt[j][0]-pt[x][0])*(pt[j][0]-pt[x][0]));
c=sqrt((pt[y][1]-pt[j][1])*(pt[y][1]-pt[j][1])+(pt[y][0]-pt[j][0])*(pt[y][0]-pt[j][0]));
e=(a+b+c)/2;
s=sqrt(e*(e-a)*(e-b)*(e-c));
h[j]=(2*s)/a;
if(h[j]>max)
{max=h[j];
z=j;
}
acutPrintf("\nh[%d]=%lf",j,h[j]);
}
acutPrintf("\n第%d点距离最大为%lf:",z+1,max);
return z;
}
在顶点遍历中可能需要采用递归和循环思路，可是我花了好些时间也不能理顺思路，真是急啊，恳请各位高手指点一下有什么好的思路和方法啊，如何去用递归和循环来实现顶点遍历的问题，在下感激不尽！！！！！