poj 3525 Most Distant Point from the Sea

题意：转换过来就是求凸多边形内的最大圆半径

思路：二分半径枚举答案，用半平面交来cheak

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cstring>
#include<cmath>
#include<algorithm>
#define inf 100000000
using namespace std;
const double pi=acos(-1.0);
const double eps=1e-8;
int n,ln,dq[100000],top,bot;
struct P{
double x,y;
P (){}
P(double x,double y):x(x),y(y){
}
P operator +(P p){
return P(x+p.x,y+p.y);
}
P operator -(P p){
return P(x-p.x,y-p.y);
}
P operator *(double d){
return P(x*d,y*d);
}
double det(P p){
return x*p.y-y*p.x;
}
};
P p[10000000];
struct Line
{
P a,b;
double angle;
};
Line l[100000],l1[100000];
P in(Line a,Line b){
return a.a+(a.b-a.a)*((b.b-b.a).det(b.a-a.a) / (b.b-b.a).det(a.b-a.a));
}
int dblcmp(double k)
{
if(fabs(k)<eps)
return 0;
return k>0?1:-1;
}
bool cmp(Line a,Line b){
int d=dblcmp(a.angle-b.angle);
if(!d)
return dblcmp((b.a-a.a).det(b.b-a.a))<0;
return d<0;
}
bool judge(Line a,Line b,Line c){
P r=in(b,c);
return dblcmp((a.a-r).det(a.b-r))<0;
}
double dis(P a,P b){
double c=(a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y);
return sqrt(c);
}
//向量向内侧平移r距离函数
void getmove(double r)
{
for(int i=0;i<ln;i++)
{
double x=l[i].a.x-l[i].b.x;
double y=l[i].a.y-l[i].b.y;
double L=dis(l[i].a,l[i].b);
l1[i].a.x=(r*y/L+l[i].a.x);
l1[i].a.y=(l[i].a.y-r*x/L);
l1[i].b.x=(r*y/L+l[i].b.x);
l1[i].b.y=(l[i].b.y-r*x/L);
l1[i].angle=l[i].angle;
}
}
bool check(double r)
{
int len;
int i,j;
getmove(r);
for(i=0,j=0;i<ln;i++)
{
if(dblcmp(l1[i].angle-l1[j].angle)>0)
l1[++j]=l1[i];
}
len=j+1;
top=1;
bot=0;
dq[0]=0;
dq[1]=1;
for(i=2;i<len;i++)
{
while(top>bot && judge(l1[i],l1[dq[top]],l1[dq[top-1]])) top--;
while(top>bot && judge(l1[i],l1[dq[bot]],l1[dq[bot+1]])) bot++;
dq[++top]=i;
}
while(top>bot && judge(l1[dq[bot]],l1[dq[top]],l1[dq[top-1]])) top--;
while(top>bot && judge(l1[dq[top]],l1[dq[bot]],l1[dq[bot+1]])) bot++;
if(top-bot>1)
return true;
return false;
}
int main()
{
while(~scanf("%d",&n) && n)
{
ln=0;
for(int i=0;i<n;i++)
scanf("%lf%lf",&p[i].x,&p[i].y);
for(int i=0;i<n-1;i++)
{
l[ln].a=p[i];
l[ln].b=p[i+1];
l[ln].angle=atan2(l[ln].b.y-l[ln].a.y,l[ln].b.x-l[ln].a.x);
ln++;
}
l[ln].a=p[n-1];
l[ln].b=p[0];
l[ln].angle=atan2(l[ln].b.y-l[ln].a.y,l[ln].b.x-l[ln].a.x);
ln++;
sort(l,l+ln,cmp);
double l=0,lr=20000.0,m;
while(fabs(l-lr)>eps)
{
m=(l+lr)/2;
if(check(m))
l=m;
else
lr=m;
}
m=(l+lr)/2;
printf("%.6lf\n",m);
}
return 0;
}