POJ 3525 Most Distant Point from the Sea

http://poj.org/problem?id=3525

给出一个凸包，要求凸包内距离所有边的长度的最小值最大的是哪个

思路:二分答案，然后把凸包上的边移动这个距离，做半平面交看是否有解。

#include<cstdio>
#include<iostream>
#include<cmath>
#include<cstring>
#include<algorithm>
const double finf=1e10;
const double eps=1e-7;
const double Pi=acos(-1);
int n,tot;
struct Point{
    double x,y;
    Point(){}
    Point(double x0,double y0):x(x0),y(y0){}
}p[200005];
struct Line{
    Point s,e;
    double slop;
    Line(){}
    Line(Point s0,Point e0):s(s0),e(e0){}
}l[200005],L[200005],c[200005];
int read(){
    int t=0,f=1;char ch=getchar();
    while (ch<'0'||ch>'9'){if (ch=='-')f=-1;ch=getchar();}
    while ('0'<=ch&&ch<='9'){t=t*10+ch-'0';ch=getchar();}
    return t*f;
}
Point operator *(Point p,double x){
    return Point(p.x*x,p.y*x);
}
Point operator /(Point p,double x){
    return Point(p.x/x,p.y/x);
}
double operator *(Point p1,Point p2){
    return p1.x*p2.y-p1.y*p2.x;
}
Point operator -(Point p1,Point p2){
    return Point(p1.x-p2.x,p1.y-p2.y);
}
Point operator +(Point p1,Point p2){
    return Point(p1.x+p2.x,p1.y+p2.y);
}
double sqr(double x){
    return x*x;
}
double dis(Point p){
    return sqrt(sqr(p.x)+sqr(p.y));
}
Point e(Point p){
    double len=dis(p);
    p=p/len;
    return p;
}
Point turn(Point p,double x){
    double Sin=sin(x),Cos=cos(x);
    double X=Cos*p.x-Sin*p.y;
    double Y=Cos*p.y+Sin*p.x;
    return Point(X,Y);
}
bool cmp(Line p1,Line p2){
    if (p1.slop!=p2.slop) return p1.slop<p2.slop;
    else return (p1.e-p1.s)*(p2.e-p1.s)<=0;
}
void build(double mid){
    for (int i=1;i<=tot;i++){
     Point p=e(turn(l[i].e-l[i].s,Pi/2.0))*mid;
     L[i].s=l[i].s+p;
     L[i].e=l[i].e+p;
    }
    for (int i=1;i<=tot;i++)
     L[i].slop=l[i].slop;
    std::sort(L+1,L+1+tot,cmp);  
}
Point inter(Line p1,Line p2){
    double k1=(p2.e-p1.s)*(p1.e-p1.s);
    double k2=(p1.e-p1.s)*(p2.s-p1.s);
    double t=(k2/(k1+k2));
    double x=p2.s.x+(p2.e.x-p2.s.x)*t;
    double y=p2.s.y+(p2.e.y-p2.s.y)*t;
    return Point(x,y);
}
bool jud(Line p1,Line p2,Line p3){
    Point p=inter(p1,p2);
    return (p-p3.s)*(p3.e-p3.s)>0;
}
bool phi(){
    int cnt=1;
    for (int i=2;i<=tot;i++)
     if (L[i].slop!=L[i-1].slop) L[++cnt]=L[i];
    int lll=1,rrr=2;c[lll]=L[1];c[rrr]=L[2];
    for (int i=3;i<=cnt;i++){
        while (lll<rrr&&jud(c[rrr],c[rrr-1],L[i])) rrr--;
        while (lll<rrr&&jud(c[lll],c[lll+1],L[i])) lll++;
        c[++rrr]=L[i];
    } 
    while (lll<rrr&&jud(c[rrr],c[rrr-1],c[lll])) rrr--;
    while (lll<rrr&&jud(c[lll],c[lll+1],c[rrr])) lll++;
    if (rrr-lll+1>=3) return 1;
    else return 0;
}
bool check(double mid){
    build(mid);
    if (phi()) return 1;
    return 0;
}
int main(){
    while (scanf("%d",&n)!=EOF){
        if (n==0) return 0;
        for (int i=1;i<=n;i++)
         p[i].x=read(),p[i].y=read();
        p[n+1]=p[1];
        tot=0;
        for (int i=1;i<=n;i++)
         l[++tot]=Line(p[i],p[i+1]);
        for (int i=1;i<=tot;i++) l[i].slop=atan2(l[i].e.y-l[i].s.y,l[i].e.x-l[i].s.x); 
        double ll=0.0,rr=finf; 
        while (rr-ll>eps){
            double mid=(ll+rr)/2.0;
            if (check(mid)) ll=mid;
            else rr=mid;
        }
        printf("%.6f\n",ll);
    }
}