UVA LA3890 二分和半平面交

这题二分的思想很巧妙，加上基本的半平面交，

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<cctype>
#include<string>
#include<set>
#include<map>
#include<queue>
#include<stack>
using namespace std;
const double eps=1e-6;
int dcmp(double x)
{
    if(fabs(x)<eps) return 0;
    return x>0?1:-1;
    //return fabs(x) < eps ? 0 : (x > 0 ? 1 : -1);
}

struct point
{
  double x;
  double y;
  point(){}
  point(double x,double  y):x(x),y(y){}
  void in()
  {
      cin>>x>>y;
  }
  void out()
  {
      cout<<x<<' '<<y<<endl;
  }
  point operator + (const point &t) const
  {
      return point(x+t.x,y+t.y);
  }
  point operator - (const point &t) const
  {
      return point(x-t.x,y-t.y);
  }
  point operator * (const double &t) const
  {
      return point(x*t,y*t);
  }
};
double cross(point a,point b)
{
    return a.x*b.y-a.y*b.x;
}
double dot(point a,point b)
{
    return a.x*b.x+a.y*b.y;
}
double length(point a)
{
    return sqrt(dot(a,a));
}
point nomal(point t)
{
    double l=length(t);
    return  point(-t.y/l,t.x/l);
}
struct line
{
    point p;
    point v;
    double ang;
    line() {}
    line(point p,point v):p(p),v(v){
        ang=atan2(v.y,v.x);
    }
    bool operator < (const line &l) const
    {
        return ang<l.ang;
    }
};
bool onleft(line l,point p)
{
    return cross(l.v,p-l.p)>0;
}
point getintersection(line a,line b)
{
    point u=a.p-b.p;
    double t=cross(b.v,u)/cross(a.v,b.v);
    return a.p+a.v*t;
}
int halfplaneintersection(line *l,int n,point *poly)
{
    sort(l,l+n);

    int first,last;
    point *p=new point[n];
    line *q=new  line[n];
    q[first=last=0]=l[0];
    for(int i=1;i<n;i++)
    {
        while(first<last && !onleft(l[i],p[last-1])) last--;
        while(first<last && !onleft(l[i],p[first])) first++;
        q[++last]=l[i];
        if(fabs(cross(q[last].v,q[last-1].v))<eps)
        {
            last--;
            if(onleft(q[last],l[i].p)) q[last]=l[i];
        }
        if(first<last) p[last-1]=getintersection(q[last-1],q[last]);
    }
    while(first<last && !onleft(q[first],p[last-1])) last--;
    if(last-first<=1) return 0;
    p[last]=getintersection(q[last],q[first]);
    int m=0;
    for(int i=first;i<=last;i++) poly[m++]=p[i];
    return m;
}
point p[200],poly[200];
line l[200];
point v[200],v2[200];
int main()
{
    int n,m;
    double x,y;
    while(scanf("%d",&n)&&n)
    {
        for(int i=0;i<n;i++)
        {
           p[i].in();
        }
        for(int i=0;i<n;i++)
        {
            v[i]=p[(i+1)%n]-p[i];
            v2[i]=nomal(v[i]);
        }
    double left=0.0,right=20000.0;
    while(right-left>eps)
    {
        double mid=(right+left)/2;
        for(int i=0;i<n;i++)
            l[i]=line(p[i]+v2[i]*mid,v[i]);
      //  for(int i=0;i<n/2;i++)
     //   {
    //        v2[i].out();
      //      l[i].p.out();
     //       l[i].v.out();
     //   }
        m=halfplaneintersection(l,n,poly);
        if(!m) right=mid;
        else left=mid;

    }
    printf("%.6f\n",left);
    }
    return 0;
}