UVALive 3890 Most Distant Point from the Sea(凸包最大内接圆)

一个n个点的凸多边形，求多边形中离多边形边界最远的距离。实际上就是求凸包最大内接圆的半径。

利用半平面交求解，每次二分枚举半径d，然后将凸包每条边所代表的半平面沿其垂直单位法向量平移d，看所有平移后的半平面的交集是否为空。

#include<algorithm>#include<iostream>#include<cstring>#include<cstdlib>#include<fstream>#include<sstream>#include<bitset>#include<vector>#include<string>#include<cstdio>#include<cmath>#include<stack>#include<queue>#include<stack>#include<map>#include<set>#define FF(i, a, b) for(int i=a; i<b; i++)#define FD(i, a, b) for(int i=a; i>=b; i--)#define REP(i, n) for(int i=0; i<n; i++)#define CLR(a, b) memset(a, b, sizeof(a))#define debug puts("**debug**")#define LL long long#define PB push_back#define eps 1e-10using namespace std;struct Point{    double x, y;    Point (double x=0, double y=0):x(x), y(y) {}};typedef Point Vector;Vector operator + (Vector A, Vector B) { return Vector(A.x + B.x, A.y + B.y); }Vector operator - (Vector A, Vector B) { return Vector(A.x - B.x, A.y - B.y); }Vector operator * (Vector A, double p) { return Vector(A.x*p, A.y*p); }Vector operator / (Vector A, double p) { return Vector(A.x/p, A.y/p); }bool operator < (const Point& a, const Point& b){    return a.x < b.x || (a.x == b.x && a.y < b.y);}int dcmp(double x){    if(fabs(x) < eps) return 0;    return x < 0 ? -1 : 1;}bool operator == (const Point& a, const Point& b){    return dcmp(a.x-b.x) == 0 && dcmp(a.y-b.y) == 0;}double Dot(Vector A, Vector B) { return A.x*B.x + A.y*B.y; }double Length(Vector A) { return sqrt(Dot(A, A)); }double Angel(Vector A, Vector B) { return acos(Dot(A, B) / Length(A) / Length(B)); }double Cross(Vector A, Vector B) { return A.x*B.y - A.y*B.x; }double Area2(Vector A, Vector B, Vector C) { return Cross(B-A, C-A); }Vector Normal(Vector a) //a向量的垂直法向量{     return Vector(-a.y/Length(a), a.x/Length(a));}struct Line{    Point p;    Vector v;    double ang;    Line() {}    Line(Point p, Vector v): p(p), v(v) {ang = atan2(v.y, v.x); }    bool operator < (const Line& L) const    {        return ang < L.ang;    }};//点p在半平面的左边bool onLeft(Line L, Point p) { return Cross(L.v, p-L.p) > 0; }//直线交点Point GetIntersection(Line a, Line b){    Vector u = a.p-b.p;    double t = Cross(b.v, u) / Cross(a.v, b.v);    return a.p + a.v*t;}const int maxn = 200;Point p[maxn], poly[maxn];Line L[maxn];Vector v[maxn], v2[maxn];int n;//半平面交Point pp[maxn];Line qq[maxn];int HalfplaneIntersection(Line* L, int n, Point* poly){    sort(L, L+n);    int first, last;    qq[first=last=0] = L[0];    FF(i, 1, n)    {        while(first < last && !onLeft(L[i], pp[last-1])) last--;        while(first < last && !onLeft(L[i], pp[first])) first++;        qq[++last] = L[i];        if(fabs(Cross(qq[last].v, qq[last-1].v)) < eps)        {            last--;            if(onLeft(qq[last], L[i].p)) qq[last] = L[i];        }        if(first < last) pp[last-1] = GetIntersection(qq[last-1], qq[last]);    }    while(first < last && !onLeft(qq[first], pp[last-1])) last--;    if(last-first <= 1) return 0;    pp[last] = GetIntersection(qq[last], qq[first]);    int m = 0;    FF(i, first, last+1) poly[m++] = pp[i];    return m;}int main(){    while(scanf("%d", &n), n)    {        REP(i, n) scanf("%lf%lf", &p[i].x, &p[i].y);        REP(i, n)        {            v[i] = p[(i+1)%n]-p[i];            v2[i] = Normal(v[i]);        }        double l=0, r=20000, mid;        while(r - l > eps)        {            mid = (l+r) / 2.0;            REP(i, n) L[i] = Line(p[i]+v2[i]*mid, v[i]);            int m = HalfplaneIntersection(L, n, poly);            if(!m) r=mid; else l=mid;        }        printf("%.6f\n", l);    }    return 0;}