bzoj3053

 

k維最近點對。

要求輸出前m近點對，用priority_queue存前幾名就好，注意估價函數剪枝

 

#include<cstdio>
#include<cmath>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<queue>
#define fi first
#define se second
#define pii pair<int,int>
#define MK(a,b) make_pair((a),(b))
#define sqr(x) (x)*(x) 
using namespace std;
const int N=50005;

inline int read()
{
	int ans,f=1;char ch;
	while ((ch=getchar())<'0'||ch>'9') if (ch=='-') f=-1;ans=ch-'0';
	while ((ch=getchar())>='0'&&ch<='9') ans=ans*10+ch-'0';
	return ans*f;
}


int rt,n,k,D,m;

struct aa
{
	int l,r;
	int d[6],mi[6],mx[6];
}a[N];
void up(int u,int l)
{
	for (int i=0;i<k;i++) 
	a[u].mi[i]=min(a[u].mi[i],a[l].mi[i]),
	a[u].mx[i]=max(a[u].mx[i],a[l].mx[i]);
}
bool cmp(aa a,aa b) {return a.d[D]<b.d[D];}

int build(int l,int r,int dd)
{
	D=dd;int mid=(l+r)>>1;
	nth_element(a+l,a+mid,a+r+1,cmp);
	for (int i=0;i<k;i++)
		a[mid].mi[i]=a[mid].mx[i]=a[mid].d[i];
	if (l!=mid) a[mid].l=build(l,mid-1,(dd+1)%k),up(mid,a[mid].l);else a[mid].l=0;
	if (r!=mid) a[mid].r=build(mid+1,r,(dd+1)%k),up(mid,a[mid].r);else a[mid].r=0;
	return mid;
}

int x[6];
priority_queue<pii> ans;
int fdis(int u)
{
	int ans=0;
	for (int i=0;i<k;i++) ans+=sqr(x[i]-a[u].d[i]);
	return ans;
}
int f(int u)
{
	if (u==0) return 1e9;
	int ans=0;
	for (int i=0;i<k;i++)
	{
		if (a[u].mi[i]>x[i]) ans+=sqr(a[u].mi[i]-x[i]);
		if (a[u].mx[i]<x[i]) ans+=sqr(x[i]-a[u].mx[i]);
	}
	return ans;
}
void query(int u)
{
	int dis=fdis(u);
	int dl=f(a[u].l),dr=f(a[u].r);
	
	if (ans.size()<m) ans.push(MK(dis,u));
	else if (ans.top().fi>dis) {ans.pop();ans.push(MK(dis,u));}
	
	if (dl<dr)
	{if (ans.size()<m||ans.top().fi>dl) query(a[u].l);if (ans.size()<m||ans.top().fi>dr) query(a[u].r);}
	else 
	{if (ans.size()<m||ans.top().fi>dr) query(a[u].r);if (ans.size()<m||ans.top().fi>dl) query(a[u].l);}
}
int tmp[15],cnt;
void work()
{
	for (int i=1;i<=n;i++)
		for (int j=0;j<k;j++) a[i].d[j]=read();
	rt=build(1,n,0);
	int t=read();
	while (t--)
	{
		while (!ans.empty()) ans.pop();
		for (int i=0;i<k;i++) x[i]=read();m=read();
		query(rt);
		printf("the closest %d points are:\n",m);
		cnt=0;
		while (!ans.empty()) tmp[++cnt]=ans.top().se,ans.pop();
		for (int i=cnt;i>=1;i--)
		{
			for (int j=0;j<k-1;j++) printf("%d ",a[tmp[i]].d[j]);
			printf("%d\n",a[tmp[i]].d[k-1]);
		}
	}
}

int main()
{
	while (scanf("%d%d",&n,&k)!=EOF) work();
	return 0;
}