标签:print tin return turn div ota top 进入 names
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$O(n^2\log n)$。
#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; }
标签:print tin return turn div ota top 进入 names
原文地址:http://www.cnblogs.com/clrs97/p/7518579.html