The meaning of problems
Gave the n points in the plane, find the Euclidean distance to the far point k Dir
data range
First kd-tree the BST essentially only one, and then build the tree I dimension way rotation, specifically the first layer is a sort dimension, the second dimension with another sort layer, and each layer the a layer of a different dimension.
Then there kd-tree where each node has practical implications, and balanced tree like (also essentially BST).
In this title, the first large distances can be pre-k stockpiling.
Then query from a kd-tree
first need to kd-tree built out
struct node{
int x[2];
}a[maxn];
int opt;//轮换标记
bool cmp(node a,node b){
return a.x[opt]<b.x[opt];
}
void pushup(int x){
for(int i=0;i<2;i++){mx[x][i]=mn[x][i]=d[x].x[i];}//mn,mx就是当前区间的最小，最大坐标
if(lc){for(int i=0;i<2;i++){mx[x][i]=max(mx[x][i],mx[lc][i]);mn[x][i]=min(mn[x][i],mn[lc][i]);}}
if(rc){for(int i=0;i<2;i++){mx[x][i]=max(mx[x][i],mx[rc][i]);mn[x][i]=min(mn[x][i],mn[rc][i]);}}
}
void build(int &x,int L,int R){
if(L>R)return ;//这里L==R是可以的
x=++cnt;
opt=opt^1;int mid=L+R>>1;
nth_element(a+L,a+mid,a+R+1,cmp);d[x]=a[mid];//建Kd-tree就是在当前区间中选出一个最中间的
build(lc,L,mid-1);build(rc,mid+1,R);
pushup(x);
}
When the query is actually gave a coordinate, and then in the kd-tree in the investigation there any distance (current k-th largest) greater than the minimum distance of the heap, it has replaced
inline ll pf(int x){
return 1ll*x*x;
}
inline ll f(node a,node b){return pf(a.x[0]-b.x[0])+pf(a.x[1]-b.x[1]);}
inline ll g(node a,int b){return max(pf(a.x[0]-mx[b][0]),pf(a.x[0]-mn[b][0]))+max(pf(a.x[1]-mx[b][1]),pf(a.x[1]-mn[b][1]));}
inline void query(int x,node y){
ll dl=-inf,dr=-inf;
if(lc) dl=g(y,lc);//g表示一个坐标和一个节点的最大代价可能是多少
if(rc) dr=g(y,rc);
ll di=f(y,d[x]);//f表示具体的两个节点之间的举离
if(-q.top()<di){q.pop();q.push(-di);}
if(dl>dr){if(-q.top()<dl)query(lc,y);if(-q.top()<dr)query(rc,y);}//这里是剪枝如果当前第k大的距离大于子树中最大可能出现的距离，就不继续做了
else {if(-q.top()<dr)query(rc,y);if(-q.top()<dl)query(lc,y);}
}
As can be seen in a query, one kind of similar nature prune the kd-tree. Complexity is not very assured.
Total code:
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=1e5+5,inf=1<<30;
inline int read(){
char c=getchar();int t=0,f=1;
while(!isdigit(c)){if(c=='-')f=-1;c=getchar();}
while(isdigit(c)){t=(t<<3)+(t<<1)+(c^48);c=getchar();}
return t*f;
}
int n,k;
struct node{
int x[2];
}a[maxn];
int opt;
bool cmp(node a,node b){
return a.x[opt]<b.x[opt];
}
#define lc l[x]
#define rc r[x]
priority_queue<ll> q;
inline ll pf(int x){
return 1ll*x*x;
}
struct tree{
int mx[maxn<<1][2],mn[maxn<<1][2],l[maxn<<1],r[maxn<<1],cnt;
node d[maxn<<1];
void pushup(int x){
for(int i=0;i<2;i++){mx[x][i]=mn[x][i]=d[x].x[i];}
if(lc){for(int i=0;i<2;i++){mx[x][i]=max(mx[x][i],mx[lc][i]);mn[x][i]=min(mn[x][i],mn[lc][i]);}}
if(rc){for(int i=0;i<2;i++){mx[x][i]=max(mx[x][i],mx[rc][i]);mn[x][i]=min(mn[x][i],mn[rc][i]);}}
}
void build(int &x,int L,int R){
if(L>R)return ;x=++cnt;
opt=opt^1;int mid=L+R>>1;
nth_element(a+L,a+mid,a+R+1,cmp);d[x]=a[mid];
build(lc,L,mid-1);build(rc,mid+1,R);
pushup(x);
}
inline ll f(node a,node b){return pf(a.x[0]-b.x[0])+pf(a.x[1]-b.x[1]);}
inline ll g(node a,int b){return max(pf(a.x[0]-mx[b][0]),pf(a.x[0]-mn[b][0]))+max(pf(a.x[1]-mx[b][1]),pf(a.x[1]-mn[b][1]));}
inline void query(int x,node y){
ll dl=-inf,dr=-inf;
if(lc) dl=g(y,lc);
if(rc) dr=g(y,rc);
ll di=f(y,d[x]);
if(-q.top()<di){q.pop();q.push(-di);}
if(dl>dr){if(-q.top()<dl)query(lc,y);if(-q.top()<dr)query(rc,y);}
else {if(-q.top()<dr)query(rc,y);if(-q.top()<dl)query(lc,y);}
}
}t;
int main(){
//freopen("luogu4357.in","r",stdin);
//freopen("luogu4357.out","w",stdout);
n=read(),k=read();
for(int i=1;i<=n;i++){
a[i].x[0]=read();a[i].x[1]=read();
}
for(int i=1;i<=2*k;i++){
q.push(0);
}
t.build(t.l[0],1,n);
for(int i=1;i<=n;i++)t.query(1,a[i]);
printf("%lld\n",-q.top());
return 0;
}
