您的位置:首页 > 其它

【计算几何】 Codeforces Beta Round #67 (Div. 2) E. Ship's Shortest Path

2017-01-30 19:43 579 查看
读懂题意其实是模板题。就是细节略多。

#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
#define EPS 0.00000001
struct Point
{
double x,y;
Point(){}
Point(const double &X,const double &Y){x=X;y=Y;}
double Length()
{
return sqrt(x*x+y*y);
}
}p[3],a[31];
struct data
{
Point p;
int I,J;
data(){}
data(const Point &P,const int &II,const int &JJ)
{
p=P;
I=II;
J=JJ;
}
}b[31];
int e;
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 Vector &b)
{
return Vector(a.x+b.x,a.y+b.y);
}
double Cross(Vector a,Vector b)
{
return a.x*b.y-a.y*b.x;
}
Vector operator * (const double &K,const Vector &v)
{
return Vector(K*v.x,K*v.y);
}
Point GetIntersection(Point P,Vector v,Point Q,Vector w)
{
return P+(Cross(w,P-Q)/Cross(v,w))*v;
}
double dot(Vector a,Vector b)
{
return a.x*b.x+a.y*b.y;
}
bool InLine(Point a,Point P,Point Q)
{
if(fabs(Cross(a-P,Q-P))>=EPS)
return 0;
return dot(a-P,a-Q)<EPS;
}
int n;
int main()
{
//	freopen("s.in","r",stdin);
double sum=0;
for(int i=0;i<2;++i)
scanf("%lf%lf",&p[i].x,&p[i].y);
scanf("%d",&n);
for(int i=0;i<n;++i)
scanf("%lf%lf",&a[i].x,&a[i].y);
for(int i=0;i<n;++i)
if(fabs(Cross(p[1]-a[(i+1)%n],a[i]-p[0]))<EPS)
{
printf("%.9lf\n",(p[1]-p[0]).Length());
return 0;
}
for(int i=0;i<n;++i)
sum+=(a[(i+1)%n]-a[i]).Length();
/*double sum2=0;
for(int k=i+1;k<j;++k)
sum2+=(a[(k+1)%n]-a[k]).Length();
sum2+=(a[i+1]-tp).Length();
sum2+=(tp2-a[j]).Length();
double td=min((tp-p[0]).Length()+(tp2-p[1]).Length(),
(tp2-p[0]).Length()+(tp-p[1]).Length());
printf("%.9lf\n",td+min(sum-sum2,min(sum2,(tp2-tp).Length()*2.0)));*/
for(int i=0;i<n;++i)
{
Point tp=GetIntersection(p[1],p[1]-p[0],a[i],a[(i+1)%n]-a[i]);
if(InLine(tp,a[i],a[(i+1)%n]) && InLine(tp,p[0],p[1]))
{
for(int j=i+1;j<n;++j)
{
Point tp2=GetIntersection(p[1],p[1]-p[0],a[j],a[(j+1)%n]-a[j]);
if(InLine(tp2,a[j],a[(j+1)%n]) && InLine(tp2,p[0],p[1]))
b[++e]=data(tp2,i,j);
}
for(int j=1;j<=e;++j)
if(fabs(b[j].p.x-tp.x)>=EPS || fabs(b[j].p.y-tp.y)>=EPS)
{
int I=b[j].I,J=b[j].J;
Point tp2=b[j].p;
double sum2=0;
for(int k=I+1;k<J;++k)
sum2+=(a[(k+1)%n]-a[k]).Length();
sum2+=(a[I+1]-tp).Length();
sum2+=(tp2-a[J]).Length();
double td=min((tp-p[0]).Length()+(tp2-p[1]).Length(),
(tp2-p[0]).Length()+(tp-p[1]).Length());
printf("%.9lf\n",td+min(sum-sum2,min(sum2,(tp2-tp).Length()*2.0)));
return 0;
}
int I=b[1].I,J=b[1].J;
Point tp2=b[1].p;
double sum2=0;
for(int k=I+1;k<J;++k)
sum2+=(a[(k+1)%n]-a[k]).Length();
sum2+=(a[I+1]-tp).Length();
sum2+=(tp2-a[J]).Length();
double td=min((tp-p[0]).Length()+(tp2-p[1]).Length(),
(tp2-p[0]).Length()+(tp-p[1]).Length());
printf("%.9lf\n",td+min(sum-sum2,min(sum2,(tp2-tp).Length()*2.0)));
return 0;
}
}
printf("%.9lf\n",(p[1]-p[0]).Length());
return 0;
}
内容来自用户分享和网络整理,不保证内容的准确性,如有侵权内容,可联系管理员处理 点击这里给我发消息
标签: