bzoj3199 [Sdoi2013]escape

这题真tm是醉了。

就是对于每个亲戚，利用其它的亲戚对他半平面交求出其控制的范围，然后随便跑个最短路就行了

n=0卡了我一下午//////

 #include <cstdio>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 #include <cmath>
 #define N 666
 #define eps 1e-12
 using namespace std;
 int g[N][N];
 struct point{
     double x[];
     point(){}
     point(double a,double b){x[]=a,x[]=b;}
     double & operator [] (int a){return x[a];}
 }p[N];
 double dis(point a,point b){
     return sqrt((b[]-a[])*(b[]-a[])+(b[]-a[])*(b[]-a[]));
 }
 struct line{
     double a,b,c,k;
     int id;
     line(){}
     line (double x,double y,double z,int pos){
         a=x;b=y;c=z;
         k=atan2(-x,y);id=pos;
     }
     void rev(){
         a=-a;b=-b;c=-c;
         k=atan2(-a,b);
     }
 }l[N],q[N];
 int T,n,bot,tot,top,be,ans;
 double wx,wy,sx,sy,d;
 bool cmp(line a,line b){
     if(fabs(a.k-b.k)<eps)return a.c<b.c;
     return a.k<b.k;
 }
 point cross(line a,line b){
     double x=(b.c*a.b-a.c*b.b)/(a.a*b.b-a.b*b.a);
     double y=(b.c*a.a-a.c*b.a)/(a.b*b.a-a.a*b.b);
     return point(x,y);
 }
 bool judge(point a,line b){
     return a[]*b.a+a[]*b.b+b.c<-eps;
 }
 void addline (point a,point b,int id){
     point c=point((a[]+b[])/,(a[]+b[])/);
     if(a[]==b[])l[++tot]=line(0.0,1.0,-c[],id);
     else if(a[]==b[])l[++tot]=line(1.0,0.0,-c[],id);
     else l[++tot]=line(1.0,(b[]-a[])/(b[]-a[]),-c[]-(b[]-a[])/(b[]-a[])*c[],id);
     if(judge(a,l[tot]))l[tot].rev();
 }
 void work(int x){
     register int i,j;
     sort(l+,l+tot+,cmp);
     for(i=,j=;i<=tot;i++)
         if(fabs(l[i].k-l[j].k)>=eps)l[++j]=l[i];
     tot=j;
     bot=;top=;
     q[]=l[];q[]=l[];
     for(i=;i<=tot;i++){
         while(bot<top&&judge(cross(q[top-],q[top]),l[i]))top--;
         while(bot<top&&judge(cross(q[bot+],q[bot]),l[i]))bot++;
         q[++top]=l[i];
     }
     while(bot<top&&judge(cross(q[top-],q[top]),q[bot]))top--;
     for(i=bot;i<=top;i++){
         if(q[i].id)g[x][q[i].id]=g[q[i].id][x]=;
         else g[x][n+]=g[n+][x]=;
     }
 }
 int main(){
     scanf("%d",&T);
     while(T--){
         scanf("%d",&n);
         if(!n){puts("");continue;}
         memset(g,0x3f,sizeof g);
         scanf("%lf%lf%lf%lf",&wx,&wy,&sx,&sy);
         for(int i=;i<=n;i++)
             scanf("%lf%lf",&p[i][],&p[i][]);
         for(int i=;i<=n;i++){
             tot=;
             for(int j=;j<=n;j++)if(j!=i)
                 addline(p[i],p[j],j);
             addline(p[i],point(-p[i][],p[i][]),);
             addline(p[i],point(p[i][],-p[i][]),);
             addline(p[i],point(*wx-p[i][],p[i][]),);
             addline(p[i],point(p[i][],*wy-p[i][]),);
             work(i);
         }
         d=g[][];ans=;
         for(int i=;i<=n;i++){
             double now=dis(point(sx,sy),p[i]);
             if(now<d)d=now,be=i;
         }
         for(int k=;k<=n+;k++)
             for(int i=;i<=n+;i++)
                 for(int j=;j<=n+;j++)
                     g[i][j]=min(g[i][j],g[i][k]+g[k][j]);
         ans=g[be][n+]+;
         printf("%d\n",ans);
     }
     return ;
 }