對於區間[l,r]，若詢問的直線與該區間的凸殼有交，則有解。
線上段樹每個區間上維護凸殼，然後查詢時在相應凸殼上二分斜率即可，時間複雜度$O(n\log^2n)$。
#include<cstdio>
#define N 262145
typedef long long ll;
int T,n,i,c,l[N],r[N],t;
struct P{
int x,y;
P(){}
P(int _x,int _y){x=_x,y=_y;}
inline P operator-(P b){return P(x-b.x,y-b.y);}
inline ll operator*(P b){return(ll)x*b.y-(ll)y*b.x;}
}p[100010],A,B,q[1900000];
void build(int x,int a,int b){
q[l[x]=++t]=p[a];
for(int i=a+1;i<=b+1;q[++t]=p[i++])while(t>l[x]&&(p[i]-q[t])*(q[t]-q[t-1])<=0)t--;
r[x]=t;
if(a==b)return;
int mid=(a+b)>>1;
build(x<<1,a,mid),build(x<<1|1,mid+1,b);
}
inline bool ex(int l,int r){
int mid,t=l++;
while(l<=r){
mid=(l+r)>>1;
if((q[mid]-A)*(B-q[mid])<(q[mid-1]-A)*(B-q[mid-1]))l=(t=mid)+1;else r=mid-1;
}
return(q[t]-A)*(B-q[t])<0;
}
int ask(int x,int a,int b){
if(c<=a){
if(!ex(l[x],r[x]))return 0;
if(a==b)return a;
}
int mid=(a+b)>>1;
if(c<=mid){
int t=ask(x<<1,a,mid);
if(t)return t;
}
return ask(x<<1|1,mid+1,b);
}
inline void read(int&a){char c;while(!(((c=getchar())>='0')&&(c<='9')));a=c-'0';while(((c=getchar())>='0')&&(c<='9'))(a*=10)+=c-'0';}
int main(){
for(read(T);T--;puts("0")){
for(read(n),i=1;i<=n;i++)read(p[i].x),read(p[i].y);
for(t=0,build(1,1,n-1),i=1;i<n-1;i++)A=p[i],B=p[i+1],c=i+1,printf("%d ",ask(1,1,n-1));
}
return 0;
}