[省选前题目整理][BZOJ 1038][ZJOI 2008]瞭望塔(半平面交)

题目链接

思路

代码

#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <algorithm>
#include <cmath>

#define MAXN 1000
#define EPS 1e-6

using namespace std;

struct Point
{
double x,y;
Point(){}
Point(double _x,double _y):x(_x),y(_y){}
}points[MAXN];

struct Line
{
Point s,t;
double k,b,ang;
Line(){}
Line(Point _s,Point _t):s(_s),t(_t)
{
k=(s.y-t.y)/(s.x-t.x);
b=s.y-k*s.x;
ang=atan2(t.y-s.y,t.x-s.x);
}
}lines[MAXN],stack[MAXN];

bool cmp(Line a,Line b)
{
if(fabs(a.ang-b.ang)<EPS) return a.b<b.b;
return a.ang<b.ang;
}

int top=0;

Point operator-(Point a,Point b)
{
return Point(a.x-b.x,a.y-b.y);
}

double cross(Point a,Point b)
{
return a.x*b.y-a.y*b.x;
}

Point getLineIntersec(Line x,Line y)
{
Point o;
o.x=(x.b-y.b)/(y.k-x.k);
o.y=x.k*o.x+x.b;
return o;
}

void HalfPanelIntersec(int n)
{
sort(lines+1,lines+n+1,cmp);
top=0;
for(int i=1;i<=n;i++)
if(i==n||fabs(lines[i].k-lines[i+1].k)>EPS)
{
while(top>=2&&getLineIntersec(lines[i],stack[top]).x<getLineIntersec(stack[top-1],stack[top]).x) top--;
stack[++top]=lines[i];
}
}

double calc1(double x)
{
double maxh=0;
for(int i=1;i<=top;i++)
maxh=max(maxh,stack[i].k*x+stack[i].b);
return maxh;
}

double calc2(int n,double x)
{
for(int i=2;i<=n;i++)
{
if(points[i].x>=x)
return points[i].y-(points[i].y-points[i-1].y)*(points[i].x-x)/(points[i].x-points[i-1].x);
}
return 0;
}

int main()
{
int n;
double ans=1e12;
scanf("%d",&n);
for(int i=1;i<=n;i++)
scanf("%lf",&points[i].x);
for(int i=1;i<=n;i++)
scanf("%lf",&points[i].y);
for(int i=1;i<n;i++)
lines[i]=Line(points[i],points[i+1]);
HalfPanelIntersec(n-1);
for(int i=1;i<=n;i++)
ans=min(ans,calc1(points[i].x)-points[i].y); //!!!!!
for(int i=1;i<top;i++)
{
Point tmp=getLineIntersec(stack[i],stack[i+1]); //!!!!
ans=min(ans,tmp.y-calc2(n,tmp.x));
}
printf("%.3lf\n",ans);
return 0;
}