题意:n个旅馆,每个有花费,m个查询,查询在某个点在c花费范围内的距离最小的旅馆
题解:kdt,建成四维,坐标两维,花费一维,id一维,实际上建树只用前两维,正常的查询,如果满足条件在更新答案即可
//#pragma GCC optimize(2)
//#pragma GCC optimize(3)
//#pragma GCC optimize(4)
//#pragma GCC optimize("unroll-loops")
//#pragma comment(linker, "/stack:200000000")
//#pragma GCC optimize("Ofast,no-stack-protector")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#include<bits/stdc++.h>
#define fi first
#define se second
#define db double
#define mp make_pair
#define pb push_back
#define pi acos(-1.0)
#define ll long long
#define vi vector<int>
#define mod 1000000007
#define ld long double
#define C 0.5772156649
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#define pll pair<ll,ll>
#define pil pair<int,ll>
#define pli pair<ll,int>
#define pii pair<int,int>
//#define cd complex<double>
#define ull unsigned long long
#define base 1000000000000000000
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
#define fin freopen("a.txt","r",stdin)
#define fout freopen("a.txt","w",stdout)
#define fio ios::sync_with_stdio(false);cin.tie(0)
template<typename T>
inline T const& MAX(T const &a,T const &b){return a>b?a:b;}
template<typename T>
inline T const& MIN(T const &a,T const &b){return a<b?a:b;}
inline void add(ll &a,ll b){a+=b;if(a>=mod)a-=mod;}
inline void sub(ll &a,ll b){a-=b;if(a<0)a+=mod;}
inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
inline ll qp(ll a,ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;a=a*a%mod,b>>=1;}return ans;}
inline ll qp(ll a,ll b,ll c){ll ans=1;while(b){if(b&1)ans=ans*a%c;a=a*a%c,b>>=1;}return ans;}
using namespace std;
const double eps=1e-8;
const ll INF=0x3f3f3f3f3f3f3f3f;
const int N=200000+10,maxn=100000+10,inf=0x3f3f3f3f;
int n,m,k,idx;
struct node{
int f[4];
bool operator <(const node&rhs)const{
return f[idx]<rhs.f[idx];
}
}a[N];
pair<ll,node>ans;
struct kdtree{
bool vis[N<<2];
node date[N<<2];
void build(int l,int r,int rt,int dep)
{
if(l>r)return ;
vis[rt]=1;
vis[rt<<1]=vis[rt<<1|1]=0;
idx=dep%k;
int m=(l+r)>>1;
nth_element(a+l,a+m,a+r+1);
date[rt]=a[m];
build(l,m-1,rt<<1,dep+1);
build(m+1,r,rt<<1|1,dep+1);
}
void query(node p,int rt,int dep)
{
if(!vis[rt])return ;
pair<ll,node>cur(0,date[rt]);
for(int i=0;i<k;i++)
cur.fi+=1ll*(cur.se.f[i]-p.f[i])*(cur.se.f[i]-p.f[i]);
int dim=dep%k;
bool fg=0;
int x=rt<<1,y=rt<<1|1;
if(p.f[dim]>=date[rt].f[dim])swap(x,y);
if(vis[x])query(p,x,dep+1);
// for(int i=0;i<3;i++)printf("%d ",date[rt].f[i]);puts("");
if(ans.fi==-1)
{
if(date[rt].f[2]<=p.f[2])ans=cur;
fg=1;
}
else
{
if(cur.fi<ans.fi&&date[rt].f[2]<=p.f[2])ans=cur;
else if(cur.fi==ans.fi&&date[rt].f[2]<=p.f[2]&&date[rt].f[3]<ans.se.f[3])ans=cur;
if(1ll*(p.f[dim]-date[rt].f[dim])*(p.f[dim]-date[rt].f[dim])<ans.fi)fg=1;
}
if(vis[y]&&fg)query(p,y,dep+1);
}
}kd;
int main()
{
k=2;
int T;scanf("%d",&T);
while(T--)
{
scanf("%d%d",&n,&m);
for(int i=0; i<n; i++)
{
for(int j=0; j<3; j++)
scanf("%d",&a[i].f[j]);
a[i].f[3]=i;
}
kd.build(0,n-1,1,0);
for(;m;m--)
{
node p;for(int j=0;j<3;j++)scanf("%d",&p.f[j]);
ans.fi=-1;
for(int i=0; i<3; i++)ans.se.f[i]=0;
kd.query(p,1,0);
// puts("------");
for(int i=0;i<2;i++)printf("%d ",ans.se.f[i]);
printf("%d
",ans.se.f[2]);
}
}
return 0;
}
/********************
1
5 1
1 4 4
2 1 2
4 5 3
5 2 1
3 3 5
3 3 4
********************/