BZOJ3053:The Closest M Points(K-D Teee)

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

Solution

还是K-D Tree模板，不过这个是真正的多维KDT，做的时候把原来的0/1扩展到多维就好了

查询m远的时候开个大根堆，当答案小于堆顶的时候就push进去，然后query内部稍微改一下

因为query的时候lans和rans忘了赋初值调了半天emmm……

Code

 #include<iostream>
 #include<cstring>
 #include<cstdio>
 #include<cmath>
 #include<queue>
 #include<algorithm>
 #define N (50000+1000)
 #define INF 1e16
 using namespace std;

 struct P
 {
     long long dis,num;
     bool operator < (const P &a) const {return dis<a.dis;}
 }po;
 long long n,k,D,t,Root,m,ans[N];
 priority_queue<P>q;

 struct Node
 {
     long long Max[],Min[],d[],lson,rson;
     bool operator < (const Node &a) const {return d[D]<a.d[D];}
 }p[N],T;

 struct KDT
 {
     Node Tree[N];
     long long sqr(long long x){return x*x;}

     void Update(long long now)
     {
         for (int i=;i<k; ++i)
         {
             long long ls=Tree[now].lson, rs=Tree[now].rson;
             Tree[now].Max[i]=Tree[now].Min[i]=Tree[now].d[i];
             if (ls)
             {
                 Tree[now].Max[i]=max(Tree[now].Max[i],Tree[ls].Max[i]);
                 Tree[now].Min[i]=min(Tree[now].Min[i],Tree[ls].Min[i]);
             }
             if (rs)
             {
                 Tree[now].Max[i]=max(Tree[now].Max[i],Tree[rs].Max[i]);
                 Tree[now].Min[i]=min(Tree[now].Min[i],Tree[rs].Min[i]);
             }
         }
     }
     long long Build(long long opt,long long l,long long r)
     {
         if (l>r) return ;
         long long mid=(l+r)>>;
         D=opt; nth_element(p+l,p+mid,p+r+);
         Tree[mid]=p[mid];
         Tree[mid].lson=Build((opt+)%k,l,mid-);
         Tree[mid].rson=Build((opt+)%k,mid+,r);
         Update(mid); return mid;
     }
     long long Get_min(long long now)
     {
         long long ans=;
         for (int i=; i<k; ++i)
         {
             if (T.d[i]>Tree[now].Max[i]) ans+=sqr(T.d[i]-Tree[now].Max[i]);
             if (T.d[i]<Tree[now].Min[i]) ans+=sqr(Tree[now].Min[i]-T.d[i]);
         }
         return ans;
     }
     void Query(int now)
     {
         long long ls=Tree[now].lson, rs=Tree[now].rson, lans=INF,rans=INF;
         if (ls) lans=Get_min(ls);
         if (rs) rans=Get_min(rs);

         long long dist=;
         for (int i=; i<k; ++i)
             dist+=sqr(Tree[now].d[i]-T.d[i]);
         po.dis=dist; po.num=now;
         if (dist<q.top().dis)
             q.pop(),q.push(po);

         if (lans<rans)
         {
             if (lans<q.top().dis) Query(ls);
             if (rans<q.top().dis) Query(rs);
         }
         else
         {
             if (rans<q.top().dis) Query(rs);
             if (lans<q.top().dis) Query(ls);
         }
     }

 }KDT;

 int main()
 {
     while (scanf("%lld%lld",&n,&k)!=EOF)
     {
         for (int i=; i<=n;++i)
             for (int j=; j<k; ++j)
                 scanf("%lld",&p[i].d[j]);
         Root=KDT.Build(,,n);

         scanf("%lld",&t);
         for (int i=; i<=t; ++i)
         {
             for (int j=; j<k; ++j)
                 scanf("%lld",&T.d[j]);
             scanf("%lld",&m);
             for (int i=; i<=m; ++i)
             {
                 po.dis=INF; po.num=;
                 q.push(po);
             }
             KDT.Query(Root);

             for (int i=; i<=m; ++i)
                 ans[i]=q.top().num,q.pop();
             printf("the closest %lld points are:\n",m);
             for (int i=m; i>=; --i)
             {
                 for (int j=; j<k; ++j)
                     printf("%lld ",p[ans[i]].d[j]);
                 printf("\n");
             }
         }
     }
 }