K-D树小结

最近几天学了一发K-Dtree，有一点理解。。

首先K-Dtree是一种算法。类似于搜索，但是如果你硬要叫它数据结构也可以。。

K-D树在形态上是一颗二叉排序树，满足左儿子权值小于根节点，根节点权值小于右儿子，由于每个K-D树节点中都有对应的点，那么怎么划分权值就成为了问题。

为了把数据分散的更好，我们可以选择对每一个维度挨个枚举然后进行划分，这时候就要用到std的一个stl了，在algorithm里，nth_element(&a[l],&a[mid],&a[r+1],cmp)，cmp函数要自己写，a是Point结构体的一个数组。

struct Point{
    ll d[2],val;
    inline ll& operator [] (int x){return d[x];}
    inline bool operator < (const Point &a)const{
        return d[now]==a.d[now]?d[now^1]<a.d[now^1]:d[now]<a.d[now];
    }
}a[MAXN];

这份代码里直接重载了<，没有写cmp函数。now是当前划分的维度，在这里划分维度的标准是挨个分，而不是按方差(见其他K-D树讲解)；

如何建树？直接上代码吧。

void build(node *&o,int l,int r,int d=0){
    if(l>r)return;
    now = d;int mid = l+r>>1;
    std::nth_element(&pt[l],&pt[mid],&pt[r+1]);
    o = new node(pt[mid]);
    build(o->ls,l,mid-1,d^1);
    build(o->rs,mid+1,r,d^1);
    o->Maintain();
}

struct node{
    node *ls,*rs;
    Point point;
    ll mn[2],mx[2],sum;

    inline void Update(node *p){
        if(!p)return;
        for(int i=0;i<=1;i++)mn[i]=min(mn[i],p->mn[i]);
        for(int i=0;i<=1;i++)mx[i]=max(mx[i],p->mx[i]);
    }
    
    inline void Maintain(){
        sum = point.val;
        if(ls)Update(ls),sum+=ls->sum;
        if(rs)Update(rs),sum+=rs->sum;
    }
}*root;

然后就是查找了。

维护每个子树所对应的矩形(最大最小x，y坐标。)

然后可以把对应的min_dis()函数当做估价函数，来进行搜索QAQ。

还是直接上代码：

#include <stdio.h>
#include <cstring>
#include <iostream>
#include <queue>
#include <algorithm>
using std::max;
using std::min;
typedef long long ll;
const ll inf = (ll)1e16;
const int MAXN = 100005;
int now,n,m,k;
 
 
template<typename _t>
inline _t read(){
    _t x=0,f=1;
    char ch=getchar();
    for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-f;
    for(;isdigit(ch);ch=getchar())x=x*10+(ch^48);
    return x*f;
}
 
struct Point{
    ll a[3];
    inline bool operator < (const Point & b)const{
        return a[now]<b.a[now]||(a[now]==b.a[now]&&a[now^1]<b.a[now^1]);
    }
    ll &operator [](int x){return a[x];}
}pt[MAXN],cmp;
 
struct res{
    ll dis,id;
    bool operator < (const res & a)const{
        return dis == a.dis? id < a.id : dis > a.dis;
    }
};
std::priority_queue<res>Q;
 
inline ll sqr(ll x){return x*x;}
inline ll dis(Point x,Point y){return sqr(x[0]-y[0])+sqr(x[1]-y[1]);}
 
struct node{
    node *ls,*rs;
    Point point;
    int mn[2],mx[2];
    node(Point &x){
        point = x;
        ls = rs = NULL;
        mn[0]=mx[0]=x[0];
        mn[1]=mx[1]=x[1];
    }
    inline void Maintain(node *x){
        if(x==NULL)return;
        for(int i=0;i<=1;i++)mn[i]=min(mn[i],x->mn[i]);
        for(int i=0;i<=1;i++)mx[i]=max(mx[i],x->mx[i]);
    }
    inline ll calc_dis(){
        ll Ans = 0;
        Ans = max(Ans,dis((Point){mn[0],mn[1]},cmp));
        Ans = max(Ans,dis((Point){mn[0],mx[1]},cmp));
        Ans = max(Ans,dis((Point){mx[0],mn[1]},cmp));
        Ans = max(Ans,dis((Point){mx[0],mx[1]},cmp));
        return Ans;
    }
}*root;
 
void build(node *&o,int l,int r,int d){
    if(l>r)return;
    int mid = l+r>>1;
    now = d;std::nth_element(&pt[l],&pt[mid],&pt[r+1]);
    o = new node(pt[mid]);
    build(o->ls,l,mid-1,d^1);
    build(o->rs,mid+1,r,d^1);
    o->Maintain(o->ls);
    o->Maintain(o->rs);
}
 
inline void Query(node *rt){
    if(rt==NULL)return;
    if(Q.size()==k&&rt->calc_dis()<Q.top().dis)return;
    res ans = (res){dis(rt->point,cmp),rt->point[2]};
    if(Q.size()<k)Q.push(ans);
    else if(ans<Q.top())Q.pop(),Q.push(ans);//这个东西重载了QAQ。。
    ll dis_ls = rt->ls==NULL?inf:rt->ls->calc_dis();
    ll dis_rs = rt->rs==NULL?inf:rt->rs->calc_dis();
    if(dis_ls>dis_rs){
        Query(rt->ls);
        if(dis_rs>=Q.top().dis||Q.size()<k)Query(rt->rs);
    }
    else{
        Query(rt->rs);
        if(dis_ls>=Q.top().dis||Q.size()<k)Query(rt->ls);
    }
}
 
int main(){
    n=read<int>();
    for(int i=1;i<=n;++i)pt[i][0]=read<int>(),pt[i][1]=read<int>(),pt[i][2]=i;
    build(root,1,n,0);
    m=read<int>();
    for(int i=1;i<=m;++i){
        cmp[0]=read<int>();cmp[1]=read<int>();
        k=read<int>();
        while(!Q.empty())Q.pop();
        Query(root);
        printf("%d\n",Q.top().id);
    }
}

这是BZOJ 2626的完整代码。

题目大意：求第k远的点。

那么就维护一个堆，大小为k。。然后就可以了。

BZOJ 1941

求最远和最近的点。

用类似思路就可以了

用估价函数来判断先去哪个儿子。

#include <stdio.h>
#include <cstring>
#include <iostream>
#include <algorithm>
typedef long long ll;
const ll inf = 0x3f3f3f3f3f3f3f3fll;
const int MAXN = 500005;
using std::min;
using std::max;
int n,now;
ll Ans_min,Ans_max,Ans=inf;
 
 
template<typename _t>
inline _t read(){
    _t x=0,f=1;
    char ch=getchar();
    for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-f;
    for(;isdigit(ch);ch=getchar())x=x*10+(ch^48);
    return x*f;
}
 
struct Point{
    ll d[2];
    ll& operator [] (const int x){return d[x];}
    inline bool operator != (const Point &b)const{
        return d[0]!=b.d[0]||d[1]!=b.d[1];
    }
    bool operator < (const Point x)const{
        return d[now]<x.d[now]||(d[now]==x.d[now]&&d[now^1]<x.d[now^1]);
    }
}pt[MAXN],cur,cpy[MAXN];
 
inline ll dis(Point a,Point b){return abs(a[0]-b[0])+abs(a[1]-b[1]);} 
 
struct node{
    node *ls,*rs;
    Point point;
    ll mn[2],mx[2];
    node(Point x){
        ls=rs=NULL;
        point = x;
        mn[0]=mx[0]=x[0];
        mn[1]=mx[1]=x[1];
    }
    inline void Maintain(node *x){
        if(x==NULL)return;
        for(int i=0;i<=1;++i)mn[i]=min(mn[i],x->mn[i]);
        for(int i=0;i<=1;++i)mx[i]=max(mx[i],x->mx[i]);
    }
    inline ll min_dis(){
        ll ans = 0;
        ans += max(mn[0]-cur[0],0ll)+max(cur[0]-mx[0],0ll);
        ans += max(mn[1]-cur[1],0ll)+max(cur[1]-mx[1],0ll);
        return ans;
    }
    inline ll max_dis(){
        ll ans = 0;
        ans += max(abs(cur[0]-mn[0]),abs(cur[0]-mx[0]));
        ans += max(abs(cur[1]-mn[1]),abs(cur[1]-mx[1]));
        return ans;
    }
}*root;
 
void build(node *&o,int l,int r,int d=0){
    if(l>r)return;
    int mid = l+r>>1;now=d;
    std::nth_element(&pt[l],&pt[mid],&pt[r+1]);
    o = new node(pt[mid]);
    build(o->ls,l,mid-1,d^1);
    build(o->rs,mid+1,r,d^1);
    o->Maintain(o->ls);
    o->Maintain(o->rs);
    return ;
}
 
void Query_min(node *o){
    if(o==NULL)return;
    if(o->point!=cur)Ans_min=min(Ans_min,dis(cur,o->point));
    ll dis_l = o->ls?o->ls->min_dis():inf;
    ll dis_r = o->rs?o->rs->min_dis():inf;
    if(dis_l<dis_r){
        if(o->ls)Query_min(o->ls);
        if(dis_r<=Ans_min&&o->rs)Query_min(o->rs);
    }
    else{
        if(o->rs)Query_min(o->rs);
        if(dis_l<=Ans_min&&o->ls)Query_min(o->ls);
    }
}
 
void Query_max(node *o){
    if(o==NULL)return;
    if(o->point!=cur)Ans_max=max(Ans_max,dis(cur,o->point));
    ll dis_l = o->ls?o->ls->max_dis():inf;
    ll dis_r = o->rs?o->rs->max_dis():inf;
    if(dis_l>dis_r){
        if(o->ls)Query_max(o->ls);
        if(dis_r>=Ans_max&&o->rs)Query_max(o->rs);
    }
    else{
        if(o->rs)Query_max(o->rs);
        if(dis_l>=Ans_max&&o->ls)Query_max(o->ls);
    }
}
 
inline ll Query_max(Point p){
    Ans_max = -inf;cur = p;
    Query_max(root);
    return Ans_max;
}
 
inline ll Query_min(Point p){
    Ans_min=inf;cur=p;
    Query_min(root);
    return Ans_min;
}
 
int main(){
    n=read<int>();
    for(int i=1;i<=n;i++){
        pt[i][0]=read<int>();
        pt[i][1]=read<int>();
        cpy[i]=pt[i];
    }
    build(root,1,n);
    for(int i=1;i<=n;i++)Ans = min(Query_max(cpy[i])-Query_min(cpy[i]),Ans);
    printf("%lld\n",Ans);
}

BZOJ 4520 和2626基本差不多。对总体维护一个大小为2*k的堆就行了，因为每个点都被算了两遍，所以要2*k。

#include <stdio.h>
#include <cstring>
#include <iostream>
#include <queue>
#include <algorithm>
using namespace std;
typedef long long ll;
const ll inf = (ll)1e16;
const int MAXN = 100005;
int now,n,m,k;
  
  
template<typename _t>
inline _t read(){
    _t x=0,f=1;
    char ch=getchar();
    for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-f;
    for(;isdigit(ch);ch=getchar())x=x*10+(ch^48);
    return x*f;
}
  
struct Point{
    ll a[2];
    inline bool operator < (const Point & b)const{
        return a[now]<b.a[now]||(a[now]==b.a[now]&&a[now^1]<b.a[now^1]);
    }
    ll &operator [](int x){return a[x];}
}pt[MAXN],cmp;
priority_queue<ll,vector<ll> , greater<ll> >Q;
  
inline ll sqr(ll x){return x*x;}
inline ll dis(Point x,Point y){return sqr(x[0]-y[0])+sqr(x[1]-y[1]);}
  
struct node{
    node *ls,*rs;
    Point point;
    int mn[2],mx[2];
    node(Point &x){
        point = x;
        ls = rs = NULL;
        mn[0]=mx[0]=x[0];
        mn[1]=mx[1]=x[1];
    }
    inline void Maintain(node *x){
        if(x==NULL)return;
        for(int i=0;i<=1;i++)mn[i]=min(mn[i],x->mn[i]);
        for(int i=0;i<=1;i++)mx[i]=max(mx[i],x->mx[i]);
    }
    inline ll calc_dis(){
        ll Ans = 0;
        Ans = max(Ans,dis((Point){mn[0],mn[1]},cmp));
        Ans = max(Ans,dis((Point){mn[0],mx[1]},cmp));
        Ans = max(Ans,dis((Point){mx[0],mn[1]},cmp));
        Ans = max(Ans,dis((Point){mx[0],mx[1]},cmp));
        return Ans;
    }
}*root;
  
void build(node *&o,int l,int r,int d){
    if(l>r)return;
    int mid = l+r>>1;
    now = d;nth_element(&pt[l],&pt[mid],&pt[r+1]);
    o = new node(pt[mid]);
    build(o->ls,l,mid-1,d^1);
    build(o->rs,mid+1,r,d^1);
    o->Maintain(o->ls);
    o->Maintain(o->rs);
}
  
inline void Query(node *rt){
    if(rt==NULL)return;
    if(Q.size()==k&&rt->calc_dis()<Q.top())return;
    ll ans = dis(rt->point,cmp);
    if(Q.size()<k)Q.push(ans);
    else if(ans>Q.top())Q.pop(),Q.push(ans);
    ll dis_ls = rt->ls==NULL?inf:rt->ls->calc_dis();
    ll dis_rs = rt->rs==NULL?inf:rt->rs->calc_dis();
    if(dis_ls>dis_rs){
        Query(rt->ls);
        if(dis_rs>=Q.top()||Q.size()<k)Query(rt->rs);
    }
    else{
        Query(rt->rs);
        if(dis_ls>=Q.top()||Q.size()<k)Query(rt->ls);
    }
}
  
int main(){
    n=read<int>();k=read<int>();k<<=1;
    for(int i=1;i<=n;++i)pt[i][0]=read<int>(),pt[i][1]=read<int>();
    build(root,1,n,0);
    for(int i=1;i<=n;i++)cmp=pt[i],Query(root);
    printf("%lld\n",Q.top());
}

其余题目：

bzoj2989  带插入K-D树+替罪羊思想。

bzoj4066 和2989思路差不多。

bzoj2850 巧克力王国