poj3525 Most Distant Point from the Sea
题目描述:
题解:
二分答案+半平面交。
半径范围在0到5000之间二分,每次取$mid$然后平移所有直线,判断半平面交面积是否为零。
我的eps值取的是$10^{-12}$,36ms,而且和样例一样。
(大力推荐)
代码:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include<cmath> #include<vector> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; const int N = 150; const double eps = 1e-12; int dcmp(double x) { if(fabs(x)<=eps)return 0; return x>0?1:-1; } struct Point { double x,y; Point(){} Point(double x,double y):x(x),y(y){} Point operator + (const Point&a)const{return Point(x+a.x,y+a.y);} Point operator - (const Point&a)const{return Point(x-a.x,y-a.y);} Point operator * (const double&a)const{return Point(x*a,y*a);} Point operator / (const double&a)const{return Point(x/a,y/a);} double operator * (const Point&a)const{return x*a.x+y*a.y;} double operator ^ (const Point&a)const{return x*a.y-y*a.x;} }; typedef Point Vector; typedef vector<Point> Pol; double ang(const Vector&a){return atan2(a.x,a.y);} double lth(const Vector&a){return sqrt(a*a);} Vector Vil(const Vector&a){return Vector(-a.y,a.x)/lth(a);} struct Line { Point p; Vector v; Line(){} Line(Point p,Vector v):p(p),v(v){} Line operator + (const Vector&a)const{return Line(p+a,v);} bool operator < (const Line&a)const{return ang(v)<ang(a.v);} }; int n; Point p[N],tp[N]; Vector vp[N]; Line s0[N],s[N],ts[N]; bool Onleft(Line l,Point p) { return dcmp(l.v^(p-l.p))>0; } Point L_L(Line a,Line b) { double t = ((b.p-a.p)^(b.v))/(a.v^b.v); return a.p+a.v*t; } double S_(Pol&P) { double ans = 0.0; for(int i=1,lim=(int)P.size();i<lim;i++) ans+=((P[i-1]-P[0])^(P[i]-P[0])); return fabs(ans)/2; } double bpmj() { int hd,tl; ts[hd=tl=0]=s[1]; for(int i=2;i<=n;i++) { while(hd<tl&&!Onleft(s[i],tp[tl-1]))tl--; while(hd<tl&&!Onleft(s[i],tp[hd]))hd++; ts[++tl] = s[i]; if(!dcmp(s[i].v^ts[tl-1].v)) { tl--; if(Onleft(ts[tl],s[i].p))ts[tl]=s[i]; } tp[tl-1]=L_L(ts[tl-1],ts[tl]); } while(hd<tl&&!Onleft(ts[hd],tp[tl-1]))tl--; if(tl-hd<=1)return 0; tp[tl]=L_L(ts[hd],ts[tl]); Pol P; for(int i=hd;i<=tl;i++)P.push_back(tp[i]); return S_(P); } bool check(double mid) { for(int i=1;i<=n;i++)s[i]=s0[i]+vp[i]*mid; return dcmp(bpmj())>0; } int main() { while(scanf("%d",&n)&&n!=0) { for(int i=1;i<=n;i++)scanf("%lf%lf",&p[i].x,&p[i].y); for(int i=1;i<n;i++)s0[i]=Line(p[i],p[i+1]-p[i]); s0[n]=Line(p[n],p[1]-p[n]); sort(s0+1,s0+1+n); for(int i=1;i<=n;i++) vp[i]=Vil(s0[i].v); double l = 0.0,r = 5000.0; while(dcmp(r-l)) { double mid = (l+r)/2; if(check(mid))l=mid; else r=mid; } printf("%lf\n",r); } return 0; }