最远点 决策单调性

最远点

描述:

给你一个n个点的凸多边形，求离每一个点最远的点。

题解：

容易发现随着顺时针做每一个点其最远点也会顺时针旋转

于是可以决策单调性

因为从1的最远点开始转可以转过1

所以将数组扩成两倍做

//#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,avx2,popcnt,tune=native")

//#include <immintrin.h>
//#include <emmintrin.h>
#include <bits/stdc++.h>
using namespace std;
#define rep(i,h,t) for (int i=h;i<=t;i++)
#define dep(i,t,h) for (int i=t;i>=h;i--)
#define ll long long
#define me(x) memset(x,0,sizeof(x))
#define IL inline
#define rint register int
inline ll rd(){
    ll x=0;char c=getchar();bool f=0;
    while(!isdigit(c)){if(c=='-')f=1;c=getchar();}
    while(isdigit(c)){x=(x<<1)+(x<<3)+(c^48);c=getchar();}
    return f?-x:x;
}
char ss[1<<24],*A=ss,*B=ss;
IL char gc()
{
    return A==B&&(B=(A=ss)+fread(ss,1,1<<24,stdin),A==B)?EOF:*A++;
}
template<class T>void maxa(T &x,T y)
{
    if (y>x) x=y;
}
template<class T>void mina(T &x,T y)
{
    if (y<x) x=y;
}
template<class T>void read(T &x)
{
    int f=1,c; while (c=gc(),c<48||c>57) if (c=='-') f=-1; x=(c^48);
    while(c=gc(),c>47&&c<58) x=x*10+(c^48); x*=f;
}
const int mo=1e9+7;
ll fsp(int x,int y)
{
    if (y==1) return x;
    ll ans=fsp(x,y/2);
    ans=ans*ans%mo;
    if (y%2==1) ans=ans*x%mo;
    return ans;
}
struct cp {
    ll x,y;
    cp operator +(cp B)
    {
        return (cp){x+B.x,y+B.y};
    }
    cp operator -(cp B)
    {
        return (cp){x-B.x,y-B.y};
    }
    ll operator *(cp B)
    {
        return x*B.y-y*B.x;
    }
    int half() { return y < 0 || (y == 0 && x < 0); }
};
const int N=2e5;
struct re{
    ll a,b,id;
}a[N];
int n,f[N];
ll d(re a,re b)   
{   
  return 1ll*(a.a-b.a)*(a.a-b.a)+1ll*(a.b-b.b)*(a.b-b.b); 
}
bool pd(int now,int pos1,int pos2)
{
    ll t1=d(a[now],a[pos1]),t2=d(a[now],a[pos2]);
    if (pos1<now||pos1>now+n)   t1=-t1;
    if (pos2<now||pos2>now+n)   t2=-t2;
    return t1!=t2?t1<t2:a[pos1].id>a[pos2].id;
}
void solve(int l,int r,int L,int R) 
{
    if (l>r)    return;
    int mid=(l+r)>>1,Mid=L;
    for (int i=L;i<=R;i++)  if (pd(mid,Mid,i)) Mid=i;
    f[mid]=a[Mid].id;
    solve(l,mid-1,L,Mid);solve(mid+1,r,Mid,R);
}
int main()
{
   ios::sync_with_stdio(false);
   int T=rd();
   while (T--)
   {
        n=rd();
        rep(i,1,n)
        {
            a[i].a=rd(); a[i].b=rd(); a[i].id=i;
            a[i+n]=a[i];
        }
        solve(1,n,1,n*2);
        rep(i,1,n) cout<<f[i]<<endl;
   }
   return 0;
}