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//////
1 #include <cstdio> 2 #include <cstring> 3 #include <iostream> 4 #include <algorithm> 5 #include <cmath> 6 #define N 666 7 #define eps 1e-12 8 using namespace std; 9 int g[N][N]; 10 struct point{ 11 double x[2]; 12 point(){} 13 point(double a,double b){x[0]=a,x[1]=b;} 14 double & operator [] (int a){return x[a];} 15 }p[N]; 16 double dis(point a,point b){ 17 return sqrt((b[0]-a[0])*(b[0]-a[0])+(b[1]-a[1])*(b[1]-a[1])); 18 } 19 struct line{ 20 double a,b,c,k; 21 int id; 22 line(){} 23 line (double x,double y,double z,int pos){ 24 a=x;b=y;c=z; 25 k=atan2(-x,y);id=pos; 26 } 27 void rev(){ 28 a=-a;b=-b;c=-c; 29 k=atan2(-a,b); 30 } 31 }l[N],q[N]; 32 int T,n,bot,tot,top,be,ans; 33 double wx,wy,sx,sy,d; 34 bool cmp(line a,line b){ 35 if(fabs(a.k-b.k)<eps)return a.c<b.c; 36 return a.k<b.k; 37 } 38 point cross(line a,line b){ 39 double x=(b.c*a.b-a.c*b.b)/(a.a*b.b-a.b*b.a); 40 double y=(b.c*a.a-a.c*b.a)/(a.b*b.a-a.a*b.b); 41 return point(x,y); 42 } 43 bool judge(point a,line b){ 44 return a[0]*b.a+a[1]*b.b+b.c<-eps; 45 } 46 void addline (point a,point b,int id){ 47 point c=point((a[0]+b[0])/2,(a[1]+b[1])/2); 48 if(a[0]==b[0])l[++tot]=line(0.0,1.0,-c[1],id); 49 else if(a[1]==b[1])l[++tot]=line(1.0,0.0,-c[0],id); 50 else l[++tot]=line(1.0,(b[1]-a[1])/(b[0]-a[0]),-c[0]-(b[1]-a[1])/(b[0]-a[0])*c[1],id); 51 if(judge(a,l[tot]))l[tot].rev(); 52 } 53 void work(int x){ 54 register int i,j; 55 sort(l+1,l+tot+1,cmp); 56 for(i=2,j=1;i<=tot;i++) 57 if(fabs(l[i].k-l[j].k)>=eps)l[++j]=l[i]; 58 tot=j; 59 bot=1;top=2; 60 q[1]=l[1];q[2]=l[2]; 61 for(i=3;i<=tot;i++){ 62 while(bot<top&&judge(cross(q[top-1],q[top]),l[i]))top--; 63 while(bot<top&&judge(cross(q[bot+1],q[bot]),l[i]))bot++; 64 q[++top]=l[i]; 65 } 66 while(bot<top&&judge(cross(q[top-1],q[top]),q[bot]))top--; 67 for(i=bot;i<=top;i++){ 68 if(q[i].id)g[x][q[i].id]=g[q[i].id][x]=1; 69 else g[x][n+1]=g[n+1][x]=0; 70 } 71 } 72 int main(){ 73 scanf("%d",&T); 74 while(T--){ 75 scanf("%d",&n); 76 if(!n){puts("0");continue;} 77 memset(g,0x3f,sizeof g); 78 scanf("%lf%lf%lf%lf",&wx,&wy,&sx,&sy); 79 for(int i=1;i<=n;i++) 80 scanf("%lf%lf",&p[i][0],&p[i][1]); 81 for(int i=1;i<=n;i++){ 82 tot=0; 83 for(int j=1;j<=n;j++)if(j!=i) 84 addline(p[i],p[j],j); 85 addline(p[i],point(-p[i][0],p[i][1]),0); 86 addline(p[i],point(p[i][0],-p[i][1]),0); 87 addline(p[i],point(2*wx-p[i][0],p[i][1]),0); 88 addline(p[i],point(p[i][0],2*wy-p[i][1]),0); 89 work(i); 90 } 91 d=g[0][0];ans=0; 92 for(int i=1;i<=n;i++){ 93 double now=dis(point(sx,sy),p[i]); 94 if(now<d)d=now,be=i; 95 } 96 for(int k=1;k<=n+1;k++) 97 for(int i=1;i<=n+1;i++) 98 for(int j=1;j<=n+1;j++) 99 g[i][j]=min(g[i][j],g[i][k]+g[k][j]); 100 ans=g[be][n+1]+1; 101 printf("%d ",ans); 102 } 103 return 0; 104 }