求最近点对 栈溢出问题 杭电1007
我在做杭电1007的时候，先把握代码贴上来！
http://acm.hdu.edu.cn/showproblem.php?pid=1007

C/C++ code

#include<iostream.h>#include<cmath>#include<algorithm>#include<iomanip>using namespace std;#define max 100005//100000struct Point{    double x;    double y;}p[max];double Max(double a,double b){    if(a<b)        return b;    return a;}double Min(double a,double b){    if(a<b)        return a;    return b;}//Point p[max];class ClosetPoint{  private:       //保存点       int num;  public:       void Input(int iCount);//传进一个要输入的顶点数目iCount      // bool cmp_x(const Point& a,const Point& b);//比较排序      // bool cmp_y(Point a,Point b);       double Seprate(int start,int end);//分治       double Merge(double curmin,int start,int end);       double Distance(Point a,Point b);};/*bool ClosetPoint::*/cmp_x(Point a, Point b){    if(a.x!=b.x)        return a.x<b.x;    else        return a.y<b.y;}/*bool ClosetPoint::*/cmp_y(Point a,Point b){    if(a.y!=b.y)        return a.y<b.y;    else        return a.x<b.x;}void ClosetPoint::Input(int iCount)//接受输入的顶点信息{     num =iCount;   for(int i =0;i<iCount;i++)       cin>>p[i].x>>p[i].y;    sort(p,p+num,cmp_x);}double ClosetPoint::Distance(Point a,Point b){    return sqrt((b.x-a.x)*(b.x-a.x)+(b.y-a.y)*(b.y-a.y));}double ClosetPoint::Seprate(int start,int end)//分治的起始位置{    if(end-start==1)        return Distance(p[start],p[end]);    else if(end-start==2)    {        double temp1=Distance(p[start],p[start+1]);        double temp2=Distance(p[start],p[end]);        double temp3=Distance(p[start+1],p[end]);        double temp= Max(temp1,temp2);       return Max(temp,temp3);    }    int mid =(start+end)/2;    double dlm=Seprate(start,mid);    double drm=Seprate(mid+1,end);    return Min(dlm,drm);}double ClosetPoint::Merge(double curmin,int start,int end){  Point Yp[max];  int mid = (start+end)/2;  int j=0;  for(int s=0;s<num;s++)  {      if(fabs(p[s].x-p[mid].x)<=curmin)      {          Yp[j].x=p[s].x;          Yp[j++].y=p[s].y;      }  }  sort(Yp,Yp+j,cmp_y);  for(int i=0;i<j;i++)  {      for(int k=i+1;k<i+7&&k<j;k++)      {          if(Distance(Yp[i],Yp[k])<curmin)          {              curmin = Distance(Yp[i],Yp[k]);          }      }  }  return curmin;}int main(){    ClosetPoint cp;    int iCount;    while(cin>>iCount)    {    if(iCount==0)        break;    cp.Input(iCount);    double curmin = cp.Seprate(0,iCount-1);    double min = cp.Merge(curmin,0,iCount-1);   // cout<<fixed<<setprecision(2)<<min<<endl;    cout.precision(2);    cout.setf(ios::fixed);    cout.setf(ios::showpoint);    cout<<min/2<<endl;    }/*    Point p1,p2;    p1.x=0;    p1.y=0;    p2.x=1;    p2.y=1;    cout<<cp.Distance(p1,p2)<<endl; //test Distance method*/    return 0;}

如果我把max改为10000则一切正常，把max改为100000怎，运行时错误，报错时说栈溢出！求牛人帮忙解决一下！谢谢啊！

[解决办法]
软件运行时栈容量是固定的，如在某些函数内定义的局部变量（包括类对象）所占空间过大，执行到这些函数时，就可能出现运行时错误
解决方法：
<1>通过new或者malloc或使用API在内存堆中分配，记得不用时要销毁
<2>使用CreateFileMapping和MapViewOfFile映射内存文件的方式分配空间，同样不用时要反映射
[解决办法]
double ClosetPoint::Merge(double curmin,int start,int end)
{
Point Yp[max];

改成：
------------------->
Point* Yp = new Point[max];
记得最后别忘记delete Yp[];哦亲