kd-tree

题解：

学之前以为这个东西挺难的

看了之后发现好像挺简单的

思路上挺简单的，代码上基本和线段树差不多

hdu-2966

 代码：

#include <bits/stdc++.h>
using namespace std;
#define IL inline
#define rint register ll
#define rep(i,h,t) for(rint i=h;i<=t;i++)
#define dep(i,t,h) for(rint i=t;i>=h;i--) 
#define mid ((h+t)/2)
#define ll long long
#define me(x) memset(x,0,sizeof(x))
char ss[1<<24],*A=ss,*B=ss;
char gc()
{
  return A==B&&(B=(A=ss)+fread(ss,1,1<<24,stdin),A==B)?EOF:*A++;
}
template<class T> void read(T &x)
{
  rint 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<<3)+(x<<1)+(c^48); x*=f;
}
const ll INF=1e18;
const ll N=2e5+10;
ll n,id[N],cmp_d,ans;
ll x,y;
void umax(ll &x,ll y)
{ 
  if (x<y) x=y;
}
void umin(ll &x,ll y)
{
  if (x>y) x=y; 
}
struct re
{
  ll d[2],f;
};
bool cmp(re x,re y)
{
  return(x.d[cmp_d]<y.d[cmp_d]);
}
re p[N];
struct kd
{
  ll ls[N],rs[N],Mx[N],Nx[N],My[N],Ny[N];
  void updata(ll x)
  {
    if (ls[x])
    {
      umax(Mx[x],Mx[ls[x]]);
      umax(My[x],My[ls[x]]);
      umin(Nx[x],Nx[ls[x]]);
      umin(Ny[x],Ny[ls[x]]);
    }
    if (rs[x])
    {
      umax(Mx[x],Mx[rs[x]]);
      umax(My[x],My[rs[x]]);
      umin(Nx[x],Nx[rs[x]]);
      umin(Ny[x],Ny[rs[x]]); 
    }
  }
  ll build(ll h,ll t,ll o)
  {
    cmp_d=o; nth_element(p+h,p+mid,p+t+1,cmp);
    int x=mid; 
    id[p[x].f]=x;
    Mx[x]=Nx[x]=p[x].d[0];
    My[x]=Ny[x]=p[x].d[1];
    if (h!=x) ls[x]=build(h,mid-1,o^1); else ls[x]=0; //多组数据 
    if (t!=x) rs[x]=build(mid+1,t,o^1); else rs[x]=0;
    updata(x);
    return x;
  }
  ll dis(ll x1,ll y1,ll x2,ll y2)
  {
    ll x=x1-x2,y=y1-y2;
    return x*x+y*y; 
  }
  ll dis(ll k,ll x,ll y)
  {
    ll xx=0,yy=0;
    if (Mx[k]<x) xx=x-Mx[k];
    if (x<Nx[k]) xx=Nx[k]-x;
    if (My[k]<y) yy=y-My[k];
    if (y<Ny[k]) yy=Ny[k]-y;
    return (xx*xx+yy*yy);
  }
  void query(ll k,ll x,ll y)
  {
    ll dl=INF,dr=INF,d=dis(p[k].d[0],p[k].d[1],x,y);
    if (d) ans=min(ans,d);
    if (ls[k]) dl=dis(ls[k],x,y);
    if (rs[k]) dr=dis(rs[k],x,y);
    if (dl<dr)
    {
      if (dl<ans) query(ls[k],x,y);
      if (dr<ans) query(rs[k],x,y);
    } else
    {
      if (dr<ans) query(rs[k],x,y);
      if (dl<ans) query(ls[k],x,y);
    }
  }
}K;
int main()
{
  freopen("1.in","r",stdin);
  freopen("1.out","w",stdout);
  ll T;
  read(T);
  while (T--)
  {
//    K.clear();
    read(n);
    rep(i,1,n) read(p[i].d[0]),read(p[i].d[1]),p[i].f=i;
    ll rt=K.build(1,n,0);
    rep(i,1,n)
    {
      ans=INF;
      x=p[id[i]].d[0],y=p[id[i]].d[1];
      K.query(rt,x,y);
     // cout<<ans<<endl;
      printf("%lld\n",ans);
    }
  }
  return 0; 
}

 BZOJ2648

BZOJ3053