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;
}