BZOJ-1038 [ZJOI2008]瞭望塔

先求半平面交，然后建塔的地方肯定是在半平面交的交点上或者是在地面线段的交点上。

#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <iostream>
#include <algorithm>
#include <cctype>
#define rep(i, l, r) for(int i=l; i<=r; i++)
#define maxn 1009
#define linf 1e15
using namespace std;
inline int read()
{
	int x=0, f=1; char ch=getchar();
	while (!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
	while (isdigit(ch)) x=x*10+ch-'0', ch=getchar();
	return x*f;
}
struct P{double x, y;} p[maxn], p2[maxn];
struct line{P a, b; double ang;} g[maxn], l[maxn], a[maxn], q[maxn];
P operator - (P a, P b){return (P){a.x-b.x, a.y-b.y};}
double operator * (P a, P b){return a.x*b.y-a.y*b.x;}
bool operator < (line a, line b){if (a.ang==b.ang) return (a.b-a.a)*(b.a-a.a)>0; return a.ang<b.ang;}
bool operator < (P a, P b){return a.x<b.x;}
inline P inter(line a, line b)
{
	double k1=(b.b-a.a)*(a.b-a.a), k2=(a.b-a.a)*(b.a-a.a), t=k2/(k1+k2);
	return (P){b.a.x+t*(b.b.x-b.a.x), b.a.y+t*(b.b.y-b.a.y)};
}
inline bool jud(line a, line b, line v){return (inter(a, b)-v.a)*(v.b-v.a)>0;}



int n, m, L, R, cnt;
double ans;
int main()
{
	n=read(); m=n-1;
	rep(i, 1, n) p[i].x=read();
	rep(i, 1, n) p[i].y=read();
	rep(i, 1, m) g[i].a=p[i], g[i].b=p[i+1];
	rep(i, 1, m) l[i]=g[i], l[i].ang=atan2(l[i].b.y-l[i].a.y, l[i].b.x-l[i].a.x);
	sort(l+1, l+m+1);
	a[++cnt]=l[1]; rep(i, 2, m) a[a[cnt].ang!=l[i].ang ? ++cnt : cnt]=l[i];
	L=1, R=0, q[++R]=a[1], q[++R]=a[2];
	rep(i, 3, cnt)
	{
		while (L<R && jud(q[R-1], q[R], a[i])) R--;
		while (L<R && jud(q[L+1], q[L], a[i])) L++;
		q[++R]=a[i];
	}
	while (L<R && jud(q[R-1], q[R], q[L])) R--;
	while (L<R && jud(q[L+1], q[L], q[R])) L++;
	rep(i, L, R-1) 
	{
		p2[i]=inter(q[i], q[i+1]);
		if (p2[i].x>=p[1].x && p2[i].x<=p[m+1].x) p[++n]=p2[i];
	}
	sort(p+1, p+1+n);
	int k1=1, k2=L; ans=linf;
	rep(i, 1, n)
	{
		double x=p[i].x;
		while (k1<m && g[k1].b.x<x) k1++;
		while (k2<R && p2[k2].x<x) k2++;
		line v; v.a.x=v.b.x=x, v.a.y=1, v.b.y=2;
		ans=min(ans, inter(v, q[k2]).y-inter(v, g[k1]).y);
	}
	printf("%.3lf\n", ans);
	return 0;
}