[BZOJ1038]ZJOI2008瞭望塔|半平面交

考慮某個村莊可以被看見的區域，發現一條線段的上方就是可以看見端點的區域，那就把所有線段扔進去做半平面交，不要忘記了要加上兩條左右邊界。。求出來之後發現答案要麼是某個村莊往上到半平面交的一段距離，要麼是半平面交的某個點向下到輪廓線的一段距離，兩個都是線性函數嘛。。根據這兩種情況枚舉更新答案即可。。

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<memory.h>
#define h1 que[head]
#define h2 que[head+1]
#define t1 que[tail]
#define t2 que[tail-1]
#define N 310
#define eps 1e-8
using namespace std;
struct point{
	double x,y;
	point(){}
	point(double x,double y):x(x),y(y){}
}p[N];
struct line{
	point p,v;
	double angle;
}a[N],que[N];
int i,head,tail,cnt,n;
double x[N],y[N],ans;
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 operator*(point a,point b){return a.x*b.y-a.y*b.x;}
bool cmp(line a,line b)//被包含的放在前面 
{
	if (a.angle!=b.angle) return a.angle<b.angle;else return b.v*(a.p-b.p)>0;
}
point cross(line a,line b)
{
	point x1=a.p+a.v;
	double k1=(b.p-a.p)*b.v,k2=b.v*(b.p-x1),k=k1/(k1+k2);
	return point(k*a.v.x+a.p.x,k*a.v.y+a.p.y);
}
bool judge(line x1,line x2,line i)
{
	point t=cross(x1,x2);
//	printf("%.3lf %.3lf\n",t.x,t.y);
	return i.v*(t-i.p)>0;
}
void push(line x)
{
	while (head<tail&&!judge(t1,t2,x)) tail--;
	while (head<tail&&!judge(h1,h2,x)) head++;
	que[++tail]=x;
}
void hpi()
{
	sort(a+1,a+1+n,cmp);
	int i;cnt=0;
	a[0].angle=-10.12345;
	for (i=1;i<=n;i++)
		if (a[i].angle!=a[i-1].angle) a[++cnt]=a[i];
	head=1;tail=1;que[1]=a[1];
	for (i=2;i<=cnt;i++) push(a[i]);
	while (head<tail&&!judge(t1,t2,h1)) tail--;
	while (head<tail&&!judge(h1,h2,t1)) head++;
	cnt=0;
	for (i=head;i<tail;i++) p[++cnt]=cross(que[i],que[i+1]);
}
void getans()
{
	int i,j;
	ans=1e30;
	for (i=1;i<=cnt;i++)
		for (j=1;j<n-1;j++)
			if (x[j]<=p[i].x&&x[j+1]>=p[i].x)
			{
				line t,s;
				t.p=point(x[j],y[j]);t.v=point(x[j+1]-x[j],y[j+1]-y[j]);
				s.p=p[i];s.v=point(0,1);
				point c=cross(t,s);
				ans=min(ans,p[i].y-c.y);
			}
	for (i=1;i<n;i++)
		for (j=1;j<cnt;j++)
			if (p[j].x<=x[i]&&p[j+1].x>=x[i])
			{
				line t,s;
				t.p=p[j];t.v=p[j+1]-p[j];
				s.p=point(x[i],y[i]);s.v=point(0,1);
				point c=cross(t,s);
				ans=min(ans,c.y-y[i]);
			}
}
int main()
{
	freopen("1038.in","r",stdin);
//	freopen("my.out","w",stdout);
	scanf("%d",&n);
	for (i=1;i<=n;i++) scanf("%lf",&x[i]);
	for (i=1;i<=n;i++) scanf("%lf",&y[i]);
	x[0]=x[1];y[0]=y[1]+1;
	n++;x[n]=x[n-1];y[n]=y[n-1]+1;
	for (i=1;i<=n;i++) a[i].p=point(x[i-1],y[i-1]),a[i].v=point(x[i]-x[i-1],y[i]-y[i-1]);
	for (i=1;i<=n;i++) a[i].angle=atan2(a[i].v.y,a[i].v.x);
	hpi();
	getans();
	printf("%.3lf",ans);
}