【半平面交】bzoj1038 [ZJOI2008]瞭望塔

http://m.blog.csdn.net/blog/qpswwww/44105605

#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
#define EPS 0.0000001
#define N 311
typedef double db;
const db PI=acos(-1.0);
struct Point{db x,y;};
typedef Point Vector;
Vector operator - (const Point &a,const Point &b){return (Vector){a.x-b.x,a.y-b.y};}
Vector operator * (const Vector &a,const db &k){return (Vector){a.x*k,a.y*k};}
Vector operator + (const Vector &a,const Vector &b){return (Vector){a.x+b.x,a.y+b.y};}
db Cross(Vector a,Vector b){return a.x*b.y-a.y*b.x;}
struct Line
{
    Point p; Vector v; db ang;
    Line(){}
    Line(const Point &a,const Point &b)
      {
        v=b-a;
        p=a;
        ang=atan2(v.y,v.x);
        if(ang<0) ang+=2.0*PI; 
      }
};
bool operator < (const Line &a,const Line &b){return a.ang<b.ang;}
bool OnLeft(Line l,Point a){return Cross(l.v,a-l.p)>0;}
Point GetJiaodian(Line a,Line b){return a.p+a.v*(Cross(b.v,a.p-b.p)/Cross(a.v,b.v));}
int n;
Point ps[N];
Line q[N];
int head=1,tail=1;
Line ls[N];
void BPMJ()
{
    sort(ls+1,ls+n+1);
    q[1]=ls[1];
    for(int i=2;i<=n;++i)
      {
        while(head<tail&&(!OnLeft(ls[i],ps[tail-1]))) --tail;
        while(head<tail&&(!OnLeft(ls[i],ps[head]))) ++head;
        q[++tail]=ls[i];
        if(fabs(Cross(q[tail].v,q[tail-1].v))<EPS)
          {
            --tail;
            if(OnLeft(q[tail],ls[i].p))
              q[tail]=ls[i];
          }
        if(head<tail)
          ps[tail-1]=GetJiaodian(q[tail-1],q[tail]);
      }
    while(head<tail&&(!OnLeft(q[head],ps[tail-1]))) --tail;
    ps[tail]=GetJiaodian(q[tail],q[head]);
}
int nn;
Point a[N];
db ans=999999999999999999.0;
#define INF 10000000000.0
int main()
{
//  freopen("bzoj1038.in","r",stdin);
    scanf("%d",&nn);
    for(int i=1;i<=nn;++i) scanf("%lf",&a[i].x);
    for(int i=1;i<=nn;++i) scanf("%lf",&a[i].y);
    for(int i=1;i<nn;++i) ls[++n]=Line(a[i],a[i+1]);
    ls[++n]=Line((Point){INF,INF},(Point){-INF,INF});
    ls[++n]=Line((Point){-INF,INF},(Point){-INF,-INF});
    ls[++n]=Line((Point){-INF,-INF},(Point){INF,-INF});
    ls[++n]=Line((Point){INF,-INF},(Point){INF,INF});
    BPMJ();
    for(int i=head;i<=tail;++i)
      for(int j=1;j<nn;++j)
        if(ps[i].x>=a[j].x&&ps[i].x<=a[j+1].x)
          {
            db ty=a[j].y+(ps[i].x-a[j].x)/(a[j+1].x-a[j].x)*(a[j+1].y-a[j].y);
            ans=min(ans,ps[i].y-ty);
            break;
          }
    for(int i=1;i<=nn;++i)
      {
        for(int j=head;j<tail;++j)
          if(a[i].x>=ps[j].x&&a[i].x<=ps[j+1].x)
            {
              db ty=ps[j].y+(a[i].x-ps[j].x)/(ps[j+1].x-ps[j].x)*(ps[j+1].y-ps[j].y);
              ans=min(ans,ty-a[i].y);
            }
        if(a[i].x>=ps[tail].x&&a[i].x<=ps[head].x)
          {
            db ty=ps[tail].y+(a[i].x-ps[tail].x)/(ps[head].x-ps[tail].x)*(ps[head].y-ps[tail].y);
            ans=min(ans,ty-a[i].y);
          }
      }
    printf("%.3lf\n",ans);
    return 0;
}
——The Solution By AutSky_JadeK From UESTC 转载请注明出处:http://www.cnblogs.com/autsky-jadek/
posted @ 2015-06-18 21:47  AutSky_JadeK  阅读(94)  评论(0编辑  收藏  举报
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