NOI2011 智能车比赛

SPFA。

我们关键是要找到关键点，包括起点，终点，和相邻矩形接触线段的上端点和下端点（如图有红色圈住的点为关键点）。

我们要做的就是在这些关键点之间连边。

我们把这些关键的点拿出来:

其实就是一些竖直的线段。

除了S和T外，从左到右或者从右到左穿过线段所在的直线，必须在线段中穿过去，也就是说有个上边界和下边界。

如图是S到第4条竖直的线段的上边界l1和下边界l2。

我们先按X坐标从小到大排序，枚举边的起点，向左或者向右连边，如果遇到竖直的线段，用叉积更改上下边界即可。

构好图就直接SPFA即可。

#include<cstdio>
#include<cmath>
#include<algorithm>
#include<cstring>

using namespace std;

const int maxN=2000;
const double INF=1e15;
const double EPS=1e-9;

inline int dblcmp(double x){if (abs(x)<EPS)return 0;return x>0?1:-1;}
inline double sqr(double x){return x*x;}

struct Tpoint
  {
    double x,y;
    inline Tpoint(){}
    inline Tpoint(double _x,double _y){x=_x;y=_y;}
  };

inline double dis(Tpoint a,Tpoint b){return sqrt(sqr(a.x-b.x)+sqr(a.y-b.y));}
inline double det(Tpoint p0,Tpoint p1,Tpoint p2){return (p1.x-p0.x)*(p2.y-p0.y)-(p1.y-p0.y)*(p2.x-p0.x);}

int N;
Tpoint square[maxN+100][2];
Tpoint a[maxN+100][2];
int id[maxN+100][2],cnt;
int now,info[2*maxN+100];
struct Tedge{int v,next;double dis;}edge[2*maxN*2*maxN+1000];
double ans,v;
Tpoint S,T;
int eS,eT,idS,idT;

inline void addedge(int u,int v,double dis)
  {
    now++;
    edge[now].v=v;
    edge[now].dis=abs(dis);
    edge[now].next=info[u];
    info[u]=now;
  }

inline void solve(Tpoint s,int num,int l)
  {
        if (num!=idS && num!=idT && num%2==0){addedge(num,num+1,dis(a[l][0],a[l][1]));addedge(num+1,num,dis(a[l][0],a[l][1]));}
    Tpoint low,high,t1,t2;bool flag=0;
    for(int i=l-1;i>=1;i--)
      {
                if (!flag && (id[i][0]==idS || id[i][0]==idT))
                  {
                        addedge(num,id[i][0],dis(s,a[i][0]));addedge(id[i][0],num,dis(s,a[i][0]));
                        continue;
                  }
                if (!flag)
                  {
                        addedge(num,id[i][0],dis(s,a[i][0]));addedge(id[i][0],num,dis(s,a[i][0]));
                        addedge(num,id[i][1],dis(s,a[i][1]));addedge(id[i][1],num,dis(s,a[i][1]));
                        low=a[i][0];
                        high=a[i][1];
                        flag=1;
                        continue;
                    }
        t1=a[i][0];t2=a[i][1];
                if ( dblcmp(det(s,low,t1))<=0 && dblcmp(det(s,high,t1))>=0 ){addedge(num,id[i][0],dis(s,a[i][0]));addedge(id[i][0],num,dis(s,a[i][0]));}
        if ( dblcmp(det(s,low,t2))<=0 && dblcmp(det(s,high,t2))>=0 ){addedge(num,id[i][1],dis(s,a[i][1]));addedge(id[i][1],num,dis(s,a[i][1]));}
        if (id[i][0]!=idS && id[i][0]!=idT)
        {
        if ( dblcmp( det(s,low,t2) ) == 1 ) break;
        if ( dblcmp( det(s,high,t1)) == -1 ) break;
        if ( dblcmp( det(s,low,t1) ) == -1 ) low=t1;
        if ( dblcmp( det(s,high,t2))== 1 ) high=t2;
              }
      }
  }

int head,tail,queue[7*2*maxN+100];
bool vis[2*maxN+100];
double f[2*maxN+100];
inline double SPFA()
  {
        int S=idS,T=idT;
    for(int i=1;i<=cnt;i++) f[i]=INF;
    queue[head=tail=0]=S;
    f[S]=0.0;vis[S]=1;
    while(head<=tail)
      {
        int u=queue[(head++)%(7*2*maxN+100)],v,i;double dis;
        vis[u]=0;
        for(i=info[u],v=edge[i].v,dis=edge[i].dis;i!=-1;i=edge[i].next,v=edge[i].v,dis=edge[i].dis)
          if ( dblcmp(dis+f[u]-f[v])==-1 )
            {
              f[v]=dis+f[u];
              if (!vis[v])
                {
                  vis[v]=1;
                  queue[(++tail)%(7*2*maxN+100)]=v;
                  if ( dblcmp(f[queue[head%(7*2*maxN+100)]]-f[queue[tail%(7*2*maxN+100)]])==1 ) swap(queue[tail%(7*2*maxN+100)],queue[head%(7*2*maxN+100)]);
                }
            }
        }
      return abs(f[T]);
  }

int main()
  {
    freopen("car.in","r",stdin);
    freopen("car.out","w",stdout);
    scanf("%d\n",&N);
    for(int i=1;i<=N;i++)scanf("%lf%lf%lf%lf\n",&square[i][0].x,&square[i][0].y,&square[i][1].x,&square[i][1].y);
        scanf("%lf%lf\n",&S.x,&S.y);
    for(int i=1;i<=N;i++)
          if (dblcmp(square[i][0].x-S.x)<=0 && dblcmp(S.x-square[i][1].x)<=0 && dblcmp(square[i][0].y-S.y)<=0 && dblcmp(S.y-square[i][1].y)<=0){eS=i;break;}
    scanf("%lf%lf\n",&T.x,&T.y);
    for(int i=1;i<=N;i++)
          if (dblcmp(square[i][0].x-T.x)<=0 && dblcmp(T.x-square[i][1].x)<=0 && dblcmp(square[i][0].y-T.y)<=0 && dblcmp(T.y-square[i][1].y)<=0){eT=i;break;}
    int g=N;N=0;
    for(int i=1;i<=g;i++)
      {
                if (i==eS)
          {
                        N++;
            a[N][0].x=S.x;a[N][0].y=S.y;
            a[N][1].x=S.x;a[N][1].y=S.y;
            idS=id[N][0]=id[N][1]=++cnt;
                    }
                if (i==eT)
          {
                        N++;
            a[N][0].x=T.x;a[N][0].y=T.y;
            a[N][1].x=T.x;a[N][1].y=T.y;
            idT=id[N][0]=id[N][1]=++cnt;
                    }
                if (i==g) continue;
                N++;
        a[N][0].x=square[i][1].x;a[N][0].y=max(square[i][0].y,square[i+1][0].y);
        a[N][1].x=square[i][1].x;a[N][1].y=min(square[i][1].y,square[i+1][1].y);
        id[N][0]=++cnt;id[N][1]=++cnt;
      }
    memset(info,-1,sizeof(info));now=-1;
    for(int i=2;i<=N;i++)
          for(int j=0;j<2;j++)
              solve(a[i][j],id[i][j],i);
    ans=SPFA();
    scanf("%lf\n",&v);
    ans=ans/v;
    printf("%0.10lf\n",ans);
    return 0;
  }