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;}