[凸多边形最大内切圆][半平面交]Most Distant Point from the Sea POJ3525

The main land of Japan called Honshu is an island surrounded by the sea. In such an island, it is natural to ask a question: “Where is the most distant point from the sea?” The answer to this question for Honshu was found in 1996. The most distant point is located in former Usuda Town, Nagano Prefecture, whose distance from the sea is 114.86 km.

In this problem, you are asked to write a program which, given a map of an island, finds the most distant point from the sea in the island, and reports its distance from the sea. In order to simplify the problem, we only consider maps representable by convex polygons.

Input

The input consists of multiple datasets. Each dataset represents a map of an island, which is a convex polygon. The format of a dataset is as follows.

Every input item in a dataset is a non-negative integer. Two input items in a line are separated by a space.

n in the first line is the number of vertices of the polygon, satisfying 3 ≤ n ≤ 100. Subsequent n lines are the x- and y-coordinates of the n vertices. Line segments (xi, yi)–(xi+1, yi+1) (1 ≤ i ≤ n − 1) and the line segment (xn, yn)–(x1, y1) form the border of the polygon in counterclockwise order. That is, these line segments see the inside of the polygon in the left of their directions. All coordinate values are between 0 and 10000, inclusive.

You can assume that the polygon is simple, that is, its border never crosses or touches itself. As stated above, the given polygon is always a convex one.

The last dataset is followed by a line containing a single zero.

Output

For each dataset in the input, one line containing the distance of the most distant point from the sea should be output. An output line should not contain extra characters such as spaces. The answer should not have an error greater than 0.00001 (10−5). You may output any number of digits after the decimal point, provided that the above accuracy condition is satisfied.

Sample Input

4
0 0
10000 0
10000 10000
0 10000
3
0 0
10000 0
7000 1000
6
0 40
100 20
250 40
250 70
100 90
0 70
3
0 0
10000 10000
5000 5001
0

Sample Output

5000.000000
494.233641
34.542948
0.353553

题意： 按逆时针给出一个凸多边形，求最大的内切圆半径。

分析： 如果能想到二分半径，那这道题会容易很多。可以想象一下，如果半径为r的圆完全在多边形内部，那么各边向内平移r个距离得到的还是一个凸多边形，如果半径为r的圆不能被完全放入多边形内，那么各边向内平移r个距离构不成多边形。因此可以二分半径。对于一个确定的半径只需要判断是否可以被完全包含在凸多边形内部，具体判断的过程需要用到半平面交，显然凸多边形的半平面交为它自身，如果平移过后还是一个凸多边形，那它的半平面交一定存在，如果无法构成多边形，那半平面交也一定不存在，可以利用这个性质来写check。

具体代码如下：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
using namespace std;
const double eps = 1e-8;
const double inf = 1e20;
const double pi = acos(-1.0);
const int maxp = 1010;
int sgn(double x)
{if(fabs(x) < eps)return 0;if(x < 0)return -1;else return 1;
}
struct Point
{double x, y;Point(){}Point(double _x,double _y){x = _x, y = _y;}void input(){scanf("%lf%lf",&x,&y);}void output(){printf("%.2f %.2f\n",x,y);}bool operator == (Point b)const{return sgn(x-b.x) == 0 && sgn(y-b.y) == 0;}bool operator < (Point b)const{return sgn(x-b.x)== 0?sgn(y-b.y)<0:x<b.x;}Point operator -(const Point &b)const{return Point(x-b.x,y-b.y);}//叉积double operator ^(const Point &b)const{return x*b.y - y*b.x;}//点积double operator *(const Point &b)const{return x*b.x + y*b.y;}//返回长度double len(){return hypot(x,y);/*库函数*/}//返回长度的平方double len2(){return x*x + y*y;}//返回两点的距离double distance(Point p){return hypot(x-p.x,y-p.y);}Point operator +(const Point &b)const{return Point(x+b.x,y+b.y);}Point operator *(const double &k)const{return Point(x*k,y*k);}Point operator /(const double &k)const{return Point(x/k,y/k);}
};
struct Line
{Point s,e;Line(){}Line(Point _s,Point _e){s = _s, e = _e;}bool operator ==(Line v){return (s == v.s)&&(e == v.e);}//`根据一个点和倾斜角angle确定直线,0<=angle<pi`Line(Point p,double angle){s = p;if(sgn(angle-pi/2) == 0){e = (s + Point(0,1));}else{e = (s + Point(1,tan(angle)));}}double length(){return s.distance(e);}void input(){s.input();e.input();}bool parallel(Line v){return sgn((e-s)^(v.e-v.s)) == 0;/*两向量叉积为0*/ }Point crosspoint(Line v){double a1 = (v.e-v.s)^(s-v.s);double a2 = (v.e-v.s)^(e-v.s);return Point((s.x*a2-e.x*a1)/(a2-a1),(s.y*a2-e.y*a1)/(a2-a1));}
};
struct halfplane:public Line
{double angle;halfplane(){}//`表示向量s->e逆时针(左侧)的半平面`halfplane(Point _s,Point _e){s = _s;e = _e;}halfplane(Line v){s = v.s;e = v.e;}void calcangle(){angle = atan2(e.y-s.y,e.x-s.x);}bool operator <(const halfplane &b)const{return angle < b.angle;}
};struct halfplanes
{int n;//需要输入 halfplane hp[maxp];//需要输入，且封闭区域都在向量逆时针方向 Point p[maxp];int que[maxp];int st,ed;//队列的头尾指针，且下标从0开始，指向元素就是头和尾 void push(halfplane tmp){hp[n++] = tmp;}//去重void unique(){int m = 1;for(int i = 1;i < n;i++){if(sgn(hp[i].angle-hp[i-1].angle) != 0)hp[m++] = hp[i];//去除极角相同的情况下，位置在右边(沿向量方向)的边 else if(sgn( (hp[m-1].e-hp[m-1].s)^(hp[i].s-hp[m-1].s) ) > 0)hp[m-1] = hp[i];}n = m;}bool halfplaneinsert()//判断半平面交是否存在 {for(int i = 0;i < n;i++)hp[i].calcangle();sort(hp,hp+n);//先对倾斜角排序 unique();que[st=0] = 0;que[ed=1] = 1;p[1] = hp[0].crosspoint(hp[1]);for(int i = 2;i < n;i++){while(st<ed && sgn((hp[i].e-hp[i].s)^(p[ed]-hp[i].s))<0)ed--;while(st<ed && sgn((hp[i].e-hp[i].s)^(p[st+1]-hp[i].s))<0)st++;que[++ed] = i;if(hp[i].parallel(hp[que[ed-1]]))return false;p[ed]=hp[i].crosspoint(hp[que[ed-1]]);}while(st<ed && sgn((hp[que[st]].e-hp[que[st]].s)^(p[ed]-hp[que[st]].s))<0)ed--;while(st<ed && sgn((hp[que[ed]].e-hp[que[ed]].s)^(p[st+1]-hp[que[ed]].s))<0)st++;if(st+1>=ed)return false;//最后剩下小于三条直线，表明半平面交不存在 return true;}
};signed main()
{int n;while(~scanf("%d", &n)){if(n == 0)break;Point p[maxp];for(int i = 0; i < n; i++)p[i].input();halfplanes a;a.n = 0;for(int i = 0; i < n; i++){halfplane temp(p[i], p[(i+1)%n]);a.push(temp);}double l = 0.0, r = 5e3+300;while(r-l > eps){double mid = (l+r)/2;for(int i = 0; i < n; i++){Point vec = p[(i+1)%n]-p[i];vec = Point(-vec.y/vec.len(), vec.x/vec.len());halfplane temp(p[i]+vec*mid, p[(i+1)%n]+vec*mid);//向内平移mid长度 a.hp[i] = temp;}if(a.halfplaneinsert())l = mid;elser = mid;}printf("%.6f\n", l);}return 0;
}

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