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-10 using 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; }