pro:给定凸多边形,求凸多边形内的点到最近边界的最远距离。
sol:显然是二分一个圆,使得圆和凸多边形不相交,但是这样很难实现。 由于是凸多边形,我们可以把二分圆转化为二分凸多边形的移动。 如果每一边向左移动Mid后,任然存在“核”,则表示存在一点合法。
直线移动:移动起点即可,方向不变。
#include<bits/stdc++.h> #define ll long long #define rep(i,a,b) for(int i=a;i<=b;i++) using namespace std; const int maxn=10010; const double eps=1e-12; struct point{ double x,y; point(){} point(double xx,double yy):x(xx),y(yy){} }; struct line{ point a;//起点 point p;//起点到终点的向量 double angle; }; double dot(point a,point b){ return a.x*b.x+a.y*b.y;} double det(point a,point b){ return a.x*b.y-a.y*b.x;} point operator *(point A,double p){ return point(A.x*p,A.y*p);} point operator +(point A,point B){return point(A.x+B.x,A.y+B.y);} point operator -(point A,point B){return point(A.x-B.x,A.y-B.y);} double getangle(point a){ return atan2(a.y,a.x);} double getangle(line a){ return getangle(a.p);} point llintersect(line A,line B) { point C=A.a-B.a; double t=det(C,B.p)/det(B.p,A.p); return A.a+A.p*t; } point s[maxn]; line t[maxn],q[maxn]; int head,tail; bool cmp(line a,line b){ double A=getangle(a),B=getangle(b); point t=(b.a+b.p)-a.a; if(fabs(A-B)<eps) return det(a.p,t)>=0.0; return A<B; } bool onright(line P,line a,line b) { point o=llintersect(a,b); point Q=o-P.a; return det(Q,P.p)>0; //如果同一直线上不能相互看到,则>=0 } void Change(int i,double Mid) //(-y,x) { point p=point(-1.0*t[i].p.y,t[i].p.x); t[i].a=t[i].a+p*(Mid/sqrt(dot(t[i].p,t[i].p))); } bool halfplaneintersect(int N,double Mid) { s[N+1]=s[1]; rep(i,1,N) t[i].a=s[i],t[i].p=s[i+1]-s[i]; rep(i,1,N) Change(i,Mid); sort(t+1,t+N+1,cmp); int tot=0; rep(i,1,N-1) { if(fabs(getangle(t[i])-getangle(t[i+1]))>eps) t[++tot]=t[i]; } t[++tot]=t[N]; head=tail=0; rep(i,1,tot){ while(tail>head+1&&onright(t[i],q[tail],q[tail-1])) tail--; while(tail>head+1&&onright(t[i],q[head+1],q[head+2])) head++; q[++tail]=t[i]; } while(tail>head+1&&onright(t[head+1],q[tail],q[tail-1])) tail--;return tail-head>2; } void solve(int N) { for(int i=1;i<=N;i++) scanf("%lf%lf",&s[i].x,&s[i].y); double L=0,R=200010,Mid,ans; int T=50; while(T--){ Mid=(L+R)/2; if(halfplaneintersect(N,Mid)) ans=Mid,L=Mid; else R=Mid; } printf("%.10lf\n",ans); } int main() { int N; while(~scanf("%d",&N)&&N) solve(N); return 0; }
原文地址:https://www.cnblogs.com/hua-dong/p/10675484.html
时间: 2024-10-10 15:10:53