题目链接:http://61.187.179.132/JudgeOnline/problem.php?id=1038
题意:给出一条折线,在折线上选择一点在这点放一个竖直的杆使得杆顶能够看到折线上每个点。在这个前提下使得杆的高度最小?
思路:从折线的相邻两个点作直线。所有直线的公共区域即为杆顶的位置。杆的位置必然是公共区域的顶点(杆顶)或者折线顶点与公共区域的交点。
struct point
{
double x,y;
point(){}
point(double _x,double _y)
{
x=_x;
y=_y;
}
point operator+(point a)
{
return point(x+a.x,y+a.y);
}
point operator-(point a)
{
return point(x-a.x,y-a.y);
}
double operator*(point a)
{
return x*a.y-y*a.x;
}
point operator*(double t)
{
return point(x*t,y*t);
}
double operator^(point a)
{
return x*a.x+y*a.y;
}
point operator/(double t)
{
return point(x/t,y/t);
}
double len()
{
return sqrt(x*x+y*y);
}
point zhuanShun(double t)
{
return point(x*cos(t)+y*sin(t),y*cos(t)-x*sin(t));
}
point zhuanNi(double t)
{
return point(x*cos(t)-y*sin(t),x*sin(t)+y*cos(t));
}
point adjust(double L)
{
double d=len();
L/=d;
return point(x*L,y*L);
}
void print()
{
printf("%.3lf %.3lf\n",x+EPS,y+EPS);
}
};
point a[N],b[N],c[N];
int n,m,K;
int sgn(double x)
{
if(x>EPS) return 1;
if(x<-EPS) return -1;
return 0;
}
double cross(point a,point b,point c)
{
return (b-a)*(c-a);
}
double dot(point a,point b,point c)
{
return (b-a)^(c-a);
}
point getCross(point a,point b,point p,point q)
{
double x=(p-a)*(q-a);
double y=(q-b)*(p-b);
return (a*y+b*x)/(x+y);
}
int onseg(point a,point b,point c)
{
if(sgn(dot(a,b,c))<=0) return 1;
return 0;
}
void cut(point p,point q)
{
K=0; b[m]=b[0];
int i,x,y;
FOR0(i,m)
{
x=sgn(cross(p,q,b[i]));
y=sgn(cross(p,q,b[i+1]));
if(x>=0) c[K++]=b[i];
if(x*y<0) c[K++]=getCross(p,q,b[i],b[i+1]);
}
m=0;
for(i=0;i<K;i++) b[m++]=c[i];
}
int main()
{
RD(n);
int i,j;
FOR1(i,n) RD(a[i].x);
FOR1(i,n) RD(a[i].y);
b[0]=point(-dinf,-dinf);
b[1]=point(dinf,-dinf);
b[2]=point(dinf,dinf);
b[3]=point(-dinf,dinf);
m=4;
FOR1(i,n-1) cut(a[i],a[i+1]);
double ans=dinf;
point s,e,c;
b[m]=b[0];
FOR1(i,n)
{
s=a[i]; e=point(s.x,777);
FOR0(j,m)
{
c=getCross(s,e,b[j],b[j+1]);
if(onseg(c,b[j],b[j+1])) ans=min(ans,c.y-a[i].y);
}
}
FOR0(i,m)
{
s=b[i]; e=point(s.x,777);
FOR1(j,n-1)
{
c=getCross(s,e,a[j],a[j+1]);
if(onseg(c,a[j],a[j+1])) ans=min(ans,b[i].y-c.y);
}
}
PR(ans);
}