BZOJ2596 : [Wc2007]疯狂赛车

根据光路最快原理以及斯涅尔定律，可以得到从定点$P$进入某条直线的最佳入射角。

求出每个端点到每条线段的最佳点，建图求最短路即可。

时间复杂度$O(n^2log n)$。

#include<cstdio>
#include<cmath>
#include<algorithm>
#include<vector>
#include<queue>
using namespace std;
typedef pair<double,int>PI;
const int N=2100000,M=9100000;
const double eps=1e-6,inf=1e100;
int n,cnt,i,j,g[N],v[M],nxt[M],ed;double w[M],d[N],va,vb,si[2],co[2];
priority_queue<PI,vector<PI>,greater<PI> >q;
inline void add(int x,int y,double z){v[++ed]=y;w[ed]=z;nxt[ed]=g[x];g[x]=ed;}
inline void add2(int x,int y,double z){add(x,y,z),add(y,x,z);}
inline void ext(int x,double y){if(y+eps<d[x])q.push(PI(d[x]=y,x));}
inline int sgn(double x){
  if(x>eps)return 1;
  if(x<-eps)return -1;
  return 0;
}
struct P{
  double x,y;
  P(){}
  P(double _x,double _y){x=_x,y=_y;}
  P operator+(P b){return P(x+b.x,y+b.y);}
  P operator-(P b){return P(x-b.x,y-b.y);}
  P operator*(double b){return P(x*b,y*b);}
  P operator/(double b){return P(x/b,y/b);}
  double operator*(P b){return x*b.x+y*b.y;}
  bool operator==(P b){return !sgn(x-b.x)&&!sgn(y-b.y);}
  double len(){return hypot(x,y);}
  P rotate(double s,double c){return P(x*c-y*s,x*s+y*c);}
  P rot90(){return P(-y,x);}
}a[1010];
struct E{
  double x;int y;
  E(){}
  E(double _x,int _y){x=_x,y=_y;}
}e[2010];
inline bool cmp(const E&a,const E&b){return a.x<b.x;}
inline double cross(P a,P b){return a.x*b.y-a.y*b.x;}
inline bool point_on_segment(P p,P a,P b){
  return sgn(cross(b-a,p-a))==0&&sgn((p-a)*(p-b))<=0;
}
inline P line_intersection(P a,P b,P p,P q){
  double U=cross(p-a,q-p),D=cross(b-a,q-p);
  return a+(b-a)*(U/D);
}
inline void work(int st,int en){
  int i,j,m=2;
  P A=a[st],B=a[en],C=(B-A).rot90();
  e[1]=E(0,st),e[2]=E(C.len(),en);
  for(i=0;i<=n;i++)if(i!=st&&i!=en)for(j=0;j<2;j++){
    P D=line_intersection(A,B,a[i],a[i]+C.rotate(si[j],co[j]));
    if(D==A||D==B)continue;
    if(!point_on_segment(D,A,B))continue;
    cnt++;
    add2(i,cnt,(a[i]-D).len()/vb);
    e[++m]=E((D-A).len(),cnt);
  }
  sort(e+1,e+m+1,cmp);
  for(i=1;i<m;i++)add2(e[i].y,e[i+1].y,(e[i+1].x-e[i].x)/va);
}
int main(){
  scanf("%d%lf%lf",&n,&va,&vb);
  for(i=1;i<=n;i++)scanf("%lf%lf",&a[i].x,&a[i].y);
  if(sgn(va-vb)<=0)return printf("%.10f",(a[0]-a[n]).len()/vb),0;
  si[0]=vb/va;
  co[0]=sqrt(1.0-si[0]*si[0]);
  si[1]=-si[0];
  co[1]=co[0];
  cnt=n;
  for(i=0;i<=n;i++)for(j=0;j<i;j++)add2(i,j,(a[i]-a[j]).len()/vb);
  for(i=0;i<n;i++)work(i,i+1);
  for(i=0;i<=cnt;i++)d[i]=inf;
  ext(0,0);
  while(!q.empty()){
    PI t=q.top();q.pop();
    if(t.first-eps>d[t.second])continue;
    for(i=g[t.second];i;i=nxt[i])ext(v[i],t.first+w[i]);
  }
  return printf("%.10f",d[n]),0;
}