题意：选取多个点从第一个点到达第n个点并且要求满足条件横坐标是递增的，相邻两点之间的权值是两个点的叉积然后从第一个点到达第n个点总权值最小。
题解：因为凸包是求逆时针的时候面积包住所有点的面积最大，那么我们题目这样是顺时针，相反的就是面积最小，所以这题就是求顺时针的凸包，将中间的那些点按x从小到大，y从小到大，编号从小到大排序然后扫描法选取凸包即可，因为会有从点所以相同的点我们选取编号最小的点即可
#include<iostream>
#include<cstring>
#include<algorithm>
#include<queue>
#include<vector>
#include<cstdio>
#include<cmath>
#include<set>
#include<map>
#include<cstdlib>
#include<ctime>
#include<stack>
#include<bitset>
using namespace std;
#define mes(a,b) memset(a,b,sizeof(a))
#define rep(i,a,b) for(int i = a; i <= b; i++)
#define dec(i,a,b) for(int i = b; i >= a; i--)
#define pb push_back
#define mk make_pair
#define fi first
#define se second
#define ls rt<<1
#define rs rt<<1|1
#define lson ls,L,mid
#define rson rs,mid+1,R
#define lowbit(x) x&(-x)
typedef double db;
typedef long long int ll;
typedef pair<int,int> pii;
typedef unsigned long long ull;
const ll inf = 0x3f3f3f3f;
const int mx = 2e5+5;
const int mod = 1e9+7;
const int x_move[] = {1,-1,0,0,1,1,-1,-1};
const int y_move[] = {0,0,1,-1,1,-1,1,-1};
int n,m;
struct node{
ll x,y;
int id;
bool operator<(const node &a)const{
if(x!=a.x) return x < a.x;
if(y!=a.y) return y < a.y;
return id < a.id;
}
bool operator!=(const node &a){
return x!=a.x||y!=a.y;
}
friend node operator-(const node &a,const node &b){
node c;
c.x = a.x-b.x;
c.y = a.y-b.y;
c.id = a.id;
return c;
}
ll operator*(node &a){
return x*a.y-a.x*y;
}
}a[mx],st[mx];
int top;
bool judge(node a,node b,node c){
c = c-a;
b = b-a;
if(b*c>0)
return true;
else if(b*c==0&&c.id<b.id)
return true;
return false;
}
int main(){
//freopen("test.in","r",stdin);
//freopen("test.out","w",stdout);
int t,ca = 1;
scanf("%d",&t);
while(t--){
scanf("%d",&n);
for(int i = 1; i <= n; i++)
scanf("%lld%lld",&a[i].x,&a[i].y),a[i].id = i;
sort(a+2,a+n);
top = 0;
st[++top] = a[1];
st[++top] = a[2];
for(int i = 3; i <= n; i++)if(a[i]!=a[i-1]){
while(top>=2&&judge(st[top-1],st[top],a[i]))
top--;
st[++top] = a[i];
}
for(int i = 1; i <= top; i++)
printf("%d%c",st[i].id,i==top?'\n':' ');
}
return 0;
}