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; }