hdu 2966 In case of failure(KD-tree)

题目链接：hdu 2966 In case of failure

题意：

给你n个点，让你输出每个点到最近点的欧式距离。

题解：

KD-树裸题，板子抄的鸟神的。

#include<bits/stdc++.h>
#define F(i,a,b) for(int i=(a);i<=(b);++i)
using namespace std;
using ll=long long;

namespace KD_Tree{
    const int N=1e5+7,DI=2;
    struct Node{
        int p[DI],f,l,r,mx[DI],mi[DI];
        int operator[](const int&idx)const{return p[idx];}
        void in(int idx,int *v){f=idx;F(i,0,DI-1)p[i]=v[i];}
        void up(Node&a){
            F(i,0,DI-1){
                mi[i]=min(mi[i],a.mi[i]);
                mx[i]=max(mx[i],a.mx[i]);
            }
        }
    }T[N];
    int idx[N],cmpd,root;
    
    bool cmp(const Node&a,const Node&b){return a[cmpd]<b[cmpd];}
    void up(int x){
        if(T[x].l)T[x].up(T[T[x].l]);
        if(T[x].r)T[x].up(T[T[x].r]);
    }
    int build(int l,int r,int d=0,int f=0)
    {
        int mid=l+r>>1;
        cmpd=d%DI,nth_element(T+l,T+mid,T+r+1,cmp);
        idx[T[mid].f]=mid,T[mid].f=f;
        F(i,0,DI-1)T[mid].mi[i]=T[mid].mx[i]=T[f][i];
        T[mid].l=l!=mid?build(l,mid-1,d+1,mid):0;
        T[mid].r=r!=mid?build(mid+1,r,d+1,mid):0;
        return up(mid),mid;
    }
    ll dist(ll x,ll y=0){return x*x+y*y;}
    ll query(ll x,ll y,ll&ans,int rt=root,int d=0)
    {
        ll tmp=dist(x-T[rt][0],y-T[rt][1]);
        if(tmp)ans=min(ans,tmp);
        if(T[rt].l&&T[rt].r)
        {
            bool is=!d?(x<=T[rt][0]):(y<=T[rt][1]);
            ll dis=!d?dist(x-T[rt][0]):dist(y-T[rt][1]);
            query(x,y,ans,is?T[rt].l:T[rt].r,!d);
            if(dis<ans)query(x,y,ans,is?T[rt].r:T[rt].l,!d);
        }
        else if(T[rt].l)query(x,y,ans,T[rt].l,!d);
        else if(T[rt].r)query(x,y,ans,T[rt].r,!d);
    }
};
using namespace KD_Tree;
int t,n;

int main(){
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        F(i,1,n)
        {
            int x[2];
            scanf("%d%d",x,x+1);
            T[i].in(i,x);
        }
        root=build(1,n);
        F(i,1,n)
        {
            ll ans=1ll<<61;
            query(T[idx[i]][0],T[idx[i]][1],ans);
            printf("%lld
",ans);
        }
    }
    return 0;
}