怒艹一波KD-tree

KDtree可以快速求多维空间一个点离他最近的点。

这个题是HDU5992.裸的KD-tree

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int inf=1e9+7;
int root,n,m,key;ll len;
ll sqr(ll x)
{
    return x*x;
}
struct node
{
    int son[2],mi[3],mx[3],d[3],id;
    friend bool operator <(node a,node b)
    {
        return a.d[key]<b.d[key];
    }
    friend ll dis(node a,node b)
    {
        return sqr(a.d[1]-b.d[1])+sqr(a.d[0]-b.d[0]);
    }
}po,ans;
struct kdtree
{
    node a[200010];
    void init()
    {
        a[0].son[0]=a[0].son[1]=0;
        for (int i=0; i<3; i++)
        {
            a[0].mi[i]=inf;
            a[0].mx[i]=-inf;
        }
    }
    void update(int x)
    {
        int l=a[x].son[0],r=a[x].son[1];
        for (int i=0; i<3; i++)
        {
            a[x].mi[i]=min(a[x].d[i],min(a[l].mi[i],a[r].mi[i]));
            a[x].mx[i]=max(a[x].d[i],max(a[l].mx[i],a[r].mx[i]));
        }
    }
    int build(int l,int r,int cur)
    {
        if (l>r) return 0;
        int m=(l+r)>>1;
        key=cur; nth_element(a+l,a+m,a+r+1);
        a[m].son[0]=build(l,m-1,cur^1);
        a[m].son[1]=build(m+1,r,cur^1);
        update(m);
        return m;
    }
    void check(int q)
    {
        if(a[q].d[2]>po.d[2])return;
        ll l=dis(a[q],po);
        if((len>l)||(len==l&&a[q].id<ans.id))
        {
            ans=a[q];
            len=l;
        }
    }
    ll get(int q)
    {
        if(!q||a[q].mi[2]>po.d[2])return 1e18+5;
        ll res=0;
        for(int i=0;i<2;i++)
        {
            res+=sqr(max(0,po.d[i]-a[q].mx[i]));
            res+=sqr(max(0,a[q].mi[i]-po.d[i]));
        }
        return res;
    }
    void ask(int q)
    {
        if(a[q].mi[2]>po.d[2])return;
        check(q);
        int l=a[q].son[0],r=a[q].son[1];
        ll dl=get(l),dr=get(r);
        if(dl<dr)
        {
            if(dl<=len)ask(l);
            if(dr<=len)ask(r);
        }
        else
        {
            if(dr<=len)ask(r);
            if(dl<=len)ask(l);
        }
    }
} kd;

int main()
{
    kd.init();
    int t;scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        for (int i=1; i<=n; i++)
        {
            int x,y,z;
            scanf("%d%d%d",&x,&y,&z);
            kd.a[i].d[0]=x;
            kd.a[i].d[1]=y;
            kd.a[i].d[2]=z;
            kd.a[i].id=i;
        }
        root=kd.build(1,n,0);
        while(m--)
        {
            scanf("%d%d%d",&po.d[0],&po.d[1],&po.d[2]);
            len=1e18;kd.ask(root);
            printf("%d %d %d\n",ans.d[0],ans.d[1],ans.d[2]);
        }
    }
    return 0;
}