标签：
3053: The Closest M Points
Time Limit: 10 Sec Memory Limit: 128 MB
Submit: 442 Solved: 173
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Description
The course of Software Design and Development Practice is objectionable. ZLC is facing a serious problem .There are many points in K-dimensional space .Given a point. ZLC
need to find out the closest m points. Euclidean distance is used as the distance metric between two points. The Euclidean distance between points p and q is the length of the line segment connecting them.In Cartesian coordinates, if p = (p1, p2,..., pn) and
q = (q1, q2,..., qn) are two points in Euclidean n-space, then the distance from p to q, or from q to p is given by:
D(p,q)=D(q,p)=sqrt((q1-p1)^2+(q2-p2)^2+(q3-p3)^2…+(qn-pn)^2
Can you help him solve this problem?
软工学院的课程很讨厌！ZLC同志遇到了一个头疼的问题：在K维空间里面有许多的点，对于某些给定的点，ZLC需要找到和它最近的m个点。
（这里的距离指的是欧几里得距离：D(p, q) = D(q, p) = sqrt((q1 - p1) ^ 2 + (q2 - p2) ^ 2 + (q3 - p3) ^ 2 + ... + (qn - pn) ^ 2)
ZLC要去打Dota，所以就麻烦你帮忙解决一下了……
【Input】
第一行，两个非负整数：点数n(1 <= n <= 50000)，和维度数k(1 <= k <= 5)。
接下来的n行，每行k个整数，代表一个点的坐标。
接下来一个正整数：给定的询问数量t(1 <= t <= 10000)
下面2*t行：
第一行，k个整数：给定点的坐标
第二行：查询最近的m个点(1 <= m <= 10)
所有坐标的绝对值不超过10000。
有多组数据！
【Output】
对于每个询问，输出m+1行：
第一行："the closest m points are:" m为查询中的m
接下来m行每行代表一个点，按照从近到远排序。
保证方案唯一，下面这种情况不会出现：
2 2
1 1
3 3
1
2 2
1
Input
In the first line of the text file .there are two non-negative integers n and K. They denote respectively: the number of points, 1 <= n <= 50000, and the number of Dimensions,1
<= K <= 5. In each of the following n lines there is written k integers, representing the coordinates of a point. This followed by a line with one positive integer t, representing the number of queries,1 <= t <=10000.each query contains two lines. The k integers
in the first line represent the given point. In the second line, there is one integer m, the number of closest points you should find,1 <= m <=10. The absolute value of all the coordinates will not be more than 10000.
There are multiple test cases. Process to end of file.
Output
For each query, output m+1 lines:
The first line saying :”the closest m points are:” where m is the number of the points.
The following m lines representing m points ,in accordance with the order from near to far
It is guaranteed that the answer can only be formed in one ways. The distances from the given point to all the nearest m+1 points are different. That means input like this:
2 2
1 1
3 3
1
2 2
1
will not exist.
Sample Input
3 2
1 1
1 3
3 4
2
2 3
2
2 3
1
Sample Output
the closest 2 points are:
1 3
3 4
the closest 1 points are:
1 3
HINT
Source
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N维KD_Tree
注意用cin在Bzoj上RE
Python测数据部分，和std对拍都对
import random
import sys
#os.mknod("data.txt")
fp=open("bzoj3053_2.in",'w')
sys.stdout=fp
maxxi=10000
n=50000
k=5
print(n,k)
for i in range(0,n):
for j in range(0,k):
print(random.randint(-maxxi,maxxi+1),end=' ')
print('')
t=100
print(t)
for i in range(0,t):
for j in range(0,k):
print(random.randint(-maxxi,maxxi+1),end=' ')
print('')
print(random.randint(1,10))
fp.close()
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<queue>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define MAXN (100000+10)
#define MAXK (5)
#define MAXT (10000+10)
#define MAXXi (10000)
#define MAXM (10+10)
typedef long long ll;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return (a-b+(a-b)/F*F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
int n,K,t,m;
int cmp_d=0;
class node
{
public:
int x[MAXK];
int l,r,minv[MAXK],maxv[MAXK];
int& operator[](int i){return x[i]; }
node(){}
node(int a[])
{
memcpy(x,a,sizeof(int)*K);
l=r=0;
Rep(i,K) minv[i]=maxv[i]=x[i];
}
void read()
{
Rep(i,K) scanf("%d",&x[i]);
l=r=0;
Rep(i,K) minv[i]=maxv[i]=x[i];
}
void print()
{
printf("%d",x[0]);
For(i,K-1) printf(" %d",x[i]);
printf("\n");
}
};
priority_queue< pair<int,int> , vector< pair<int,int> > , greater<vector< pair<int,int> >::value_type> > q,qans;
int sqr(int x){return x*x;}
int dis(node a,node b){
int ans=0;
Rep(i,K) ans+=sqr(a.x[i]-b.x[i]);
return ans;
}
int dis2(node p,node a) // 点p和方形区域a的欧几里德距离
{
int ans=0;
Rep(i,K)
{
if (p.x[i]<a.minv[i]) ans+=sqr(a.minv[i]-p.x[i]);
else
if (p.x[i]>a.maxv[i]) ans+=sqr(p.x[i]-a.maxv[i]);
}
return ans;
}
int cmp(node a,node b){return a[cmp_d]<b[cmp_d]; }
class KD_Tree
{
public:
node a[MAXN];
void update(node& o)
{
if (o.l)
{
node p=a[o.l];
Rep(i,K) o.minv[i]=min(o.minv[i],p.minv[i]);
Rep(i,K) o.maxv[i]=max(o.maxv[i],p.maxv[i]);
}
if (o.r)
{
node p=a[o.r];
Rep(i,K) o.minv[i]=min(o.minv[i],p.minv[i]);
Rep(i,K) o.maxv[i]=max(o.maxv[i],p.maxv[i]);
}
}
int build(int L,int R,int nowd)
{
int m=(L+R)>>1;
cmp_d=nowd;
nth_element(a+L,a+m,a+R+1,cmp);
if (L^m) a[m].l=build(L,m-1,(nowd+1)%K);
if (R^m) a[m].r=build(m+1,R,(nowd+1)%K);
update(a[m]);
return m;
}
int root;
void _build(int L,int R,int nowd) //1-n的节点 至少为1
{
root=build(L,R,nowd);
}
node _p;
int _ans;
void ask_min_dis(int o)
{
if (o==0) return;
q.push(make_pair(dis2(_p,a[o]),o));
qans.push(make_pair(dis(_p,a[o]),o));
while (m&&!q.empty())
{
o=q.top().second;
if (o==0) return;
// cout<<q.size()<<':'<<o<<endl;
int nowdis=q.top().first;
while (m&&!qans.empty()&&nowdis>=qans.top().first)
{
a[qans.top().second].print();
qans.pop(); m--;
}
if (m==0) return ;
q.pop();
int ans1=a[o].l ? dis2(_p,a[a[o].l]) : INF; // 点p到区域内任意一点的距离的最小值
int ans2=a[o].r ? dis2(_p,a[a[o].r]) : INF;
pair<int,int> p1=make_pair(ans1,a[o].l);
pair<int,int> p2=make_pair(ans2,a[o].r);
if (ans1<ans2)
{
if(ans1<_ans) qans.push( make_pair(dis(_p,a[a[o].l]),a[o].l) ),q.push(p1);
if(ans2<_ans) qans.push( make_pair(dis(_p,a[a[o].r]),a[o].r) ),q.push(p2);
}
else {
if(ans2<_ans) qans.push( make_pair(dis(_p,a[a[o].r]),a[o].r) ),q.push(p2);
if(ans1<_ans) qans.push( make_pair(dis(_p,a[a[o].l]),a[o].l) ),q.push(p1);
}
}
while (m&&!qans.empty())
{
a[qans.top().second].print();
qans.pop(); m--;
}
}
int _ask(node p)
{
while(!q.empty()) q.pop();
while(!qans.empty()) qans.pop();
_p=p;_ans=INF;
ask_min_dis(root);
return _ans;
}
}S;
int main()
{
// freopen("bzoj3053_2.in","r",stdin);
// freopen("bzoj3053.out","w",stdout);
while(cin>>n>>K)
{
cmp_d=0;
For(i,n)
{
S.a[i].read();
}
S._build(1,n,0);
cin>>t;
For(i,t)
{
node p;
p.read();
scanf("%d",&m);
printf("the closest %d points are:\n",m);
S._ask(p);
}
}
return 0;
}
BZOJ 3053(The Closest M Points-N维KD_Tree)
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原文地址：http://blog.csdn.net/nike0good/article/details/46554525
