POJ 3525 Most Distant Point from the Sea 二分+半平面交

链接：http://poj.org/problem?id=3525

题意：给一个凸多边形的海岛，寻找海岛之中距离海边距离最长的一个点的距离。

思路：求凸多边形的最大内切圆。做法是二分半径加半平面交，将凸多边形的每条边向内部（垂直方向）收缩半径r，看每条边的半平面是否还会交出凸多边形。

P.S. 无意间找到这道题，顺便检验一下自己敲的模板。找一个好的模板真的很重要。

代码：

#include <iostream>#include <cstdio>#include <cmath>#include <algorithm>#include <vector>#include <complex>#include <queue>#define eps 1e-10#define INF 10000using namespace std;typedef complex < double > Point;typedef pair < Point , Point > Halfplane;inline double dist(Halfplane hal){    return sqrt((hal.second.imag()-hal.first.imag())*(hal.second.imag()-hal.first.imag())+(hal.second.real()-hal.first.real())*(hal.second.real()-hal.first.real()));}inline int sgn(double n){    return fabs(n)<eps?0:(n<0?-1:1);}inline double cross(Point a,Point b){    return (conj(a)*b).imag();}inline double dot(Point a,Point b){    return(conj(a)*b).real();}inline double satisfy(Point a,Halfplane p){    return sgn(cross(a-p.first,p.second-p.first))<=0;}Point crosspoint (const Halfplane &a,const Halfplane &b){    double k=cross(b.first-b.second,a.first-b.second);    k=k/(k-cross(b.first-b.second,a.second-b.second));    return a.first+(a.second-a.first)*k;}bool cmp(const Halfplane &a,const Halfplane &b){    int res= sgn(arg(a.second-a.first)-arg(b.second-b.first));    return res==0?satisfy(a.first,b):res<0;}vector < Point > halfplaneIntersection(vector<Halfplane>v){    sort(v.begin(),v.end(),cmp);    deque < Halfplane > q;    deque < Point > ans;    q.push_back(v[0]);    for(int i=1; i<int(v.size()); i++)    {        if(sgn(arg(v[i].second - v[i].first)-arg(v[i-1].second - v[i-1].first))==0)            continue;        while(ans.size()>0 && !satisfy(ans.back(),v[i]))        {            ans.pop_back();            q.pop_back();        }        while(ans.size()>0 && !satisfy(ans.front(),v[i]))        {            ans.pop_front();            q.pop_front();        }        ans.push_back(crosspoint (q.back(),v[i]));        q.push_back(v[i]);    }    while(ans.size() > 0 && !satisfy(ans.back(),q.front()))    {        ans.pop_back();        q.pop_back();    }    while(ans.size() > 0 && !satisfy(ans.front(),q.back()))    {        ans.pop_front();        q.pop_front();    }    ans.push_back(crosspoint(q.back(),q.front()));    return vector < Point > (ans.begin(),ans.end());}int main(){    int tot;    while(scanf("%d",&tot))    {        vector < Halfplane > pol;        if(tot==0)            return 0;        double fa,fb;        scanf("%lf%lf",&fa,&fb);        double la=fa,lb=fb;        for(int i=1; i<tot; i++)        {            double a,b;            scanf("%lf%lf",&a,&b);            pol.push_back (Halfplane(Point(la,lb),Point(a,b)));            la=a,lb=b;        }        pol.push_back(Halfplane(Point(la,lb),Point(fa,fb)));        double l=0,r=20000;        while(r-l>eps)        {            double mid=(l+r)/2.0;            vector < Halfplane >res;            vector < Halfplane > ::iterator it;            for(it=pol.begin(); it!=pol.end(); it++)            {                double xx=((*it).first.imag()-(*it).second.imag())/dist((*it))*mid;                double yy=((*it).second.real()-(*it).first.real())/dist((*it))*mid;                res.push_back(Halfplane(Point((*it).first.real()+xx,(*it).first.imag()+yy),Point((*it).second.real()+xx,(*it).second.imag()+yy)));            }            vector < Point > aa = halfplaneIntersection(res);            if(aa.size()<3)                r=mid;            else l=mid;        }        printf("%.6lf\n",(l+r)/2.0);    }    return 0;}