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
",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
",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
",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