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

最新推荐文章于 2019-03-12 08:22:21 发布

转载最新推荐文章于 2019-03-12 08:22:21 发布 · 84 阅读

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原文链接：http://www.cnblogs.com/refun/p/9303084.html

本文介绍了一个基于K-D Tree实现的多维空间中寻找给定点最近邻点的算法。该算法能够处理多达50,000个点，并在K维空间内找出与给定点最近的m个点。通过使用优先队列和递归查询的方式，实现了高效的最近邻搜索。

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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

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

  1 #include<iostream>
  2 #include<cstring>
  3 #include<cstdio>
  4 #include<cmath>
  5 #include<queue>
  6 #include<algorithm>
  7 #define N (50000+1000)
  8 #define INF 1e16
  9 using namespace std;
 10 
 11 struct P
 12 {
 13     long long dis,num;
 14     bool operator < (const P &a) const {return dis<a.dis;}
 15 }po;
 16 long long n,k,D,t,Root,m,ans[N];
 17 priority_queue<P>q;
 18 
 19 struct Node
 20 {
 21     long long Max[5],Min[5],d[5],lson,rson;
 22     bool operator < (const Node &a) const {return d[D]<a.d[D];}
 23 }p[N],T;
 24 
 25 struct KDT
 26 {
 27     Node Tree[N];
 28     long long sqr(long long x){return x*x;}
 29 
 30     void Update(long long now)
 31     {
 32         for (int i=0;i<k; ++i)
 33         {
 34             long long ls=Tree[now].lson, rs=Tree[now].rson;
 35             Tree[now].Max[i]=Tree[now].Min[i]=Tree[now].d[i];
 36             if (ls)
 37             {
 38                 Tree[now].Max[i]=max(Tree[now].Max[i],Tree[ls].Max[i]);
 39                 Tree[now].Min[i]=min(Tree[now].Min[i],Tree[ls].Min[i]);
 40             }
 41             if (rs)
 42             {
 43                 Tree[now].Max[i]=max(Tree[now].Max[i],Tree[rs].Max[i]);
 44                 Tree[now].Min[i]=min(Tree[now].Min[i],Tree[rs].Min[i]);
 45             }
 46         }
 47     }
 48     long long Build(long long opt,long long l,long long r)
 49     {
 50         if (l>r) return 0;
 51         long long mid=(l+r)>>1;
 52         D=opt; nth_element(p+l,p+mid,p+r+1);
 53         Tree[mid]=p[mid];
 54         Tree[mid].lson=Build((opt+1)%k,l,mid-1);
 55         Tree[mid].rson=Build((opt+1)%k,mid+1,r);
 56         Update(mid); return mid;
 57     }
 58     long long Get_min(long long now)
 59     {
 60         long long ans=0;
 61         for (int i=0; i<k; ++i)
 62         {
 63             if (T.d[i]>Tree[now].Max[i]) ans+=sqr(T.d[i]-Tree[now].Max[i]);
 64             if (T.d[i]<Tree[now].Min[i]) ans+=sqr(Tree[now].Min[i]-T.d[i]);
 65         }
 66         return ans;
 67     }
 68     void Query(int now)
 69     {
 70         long long ls=Tree[now].lson, rs=Tree[now].rson, lans=INF,rans=INF;
 71         if (ls) lans=Get_min(ls);
 72         if (rs) rans=Get_min(rs);
 73         
 74         long long dist=0;
 75         for (int i=0; i<k; ++i)
 76             dist+=sqr(Tree[now].d[i]-T.d[i]);
 77         po.dis=dist; po.num=now; 
 78         if (dist<q.top().dis)
 79             q.pop(),q.push(po);
 80             
 81         if (lans<rans)
 82         {
 83             if (lans<q.top().dis) Query(ls);
 84             if (rans<q.top().dis) Query(rs);
 85         }
 86         else
 87         {
 88             if (rans<q.top().dis) Query(rs);
 89             if (lans<q.top().dis) Query(ls);
 90         }
 91     }
 92     
 93 }KDT;
 94 
 95 int main()
 96 {
 97     while (scanf("%lld%lld",&n,&k)!=EOF)
 98     {
 99         for (int i=1; i<=n;++i)
100             for (int j=0; j<k; ++j)
101                 scanf("%lld",&p[i].d[j]);
102         Root=KDT.Build(0,1,n);
103         
104         scanf("%lld",&t);
105         for (int i=1; i<=t; ++i)
106         {
107             for (int j=0; j<k; ++j)
108                 scanf("%lld",&T.d[j]);
109             scanf("%lld",&m);
110             for (int i=1; i<=m; ++i)
111             {
112                 po.dis=INF; po.num=0;
113                 q.push(po);
114             }
115             KDT.Query(Root);
116             
117             for (int i=1; i<=m; ++i)
118                 ans[i]=q.top().num,q.pop();
119             printf("the closest %lld points are:\n",m);
120             for (int i=m; i>=1; --i)
121             {
122                 for (int j=0; j<k; ++j)
123                     printf("%lld ",p[ans[i]].d[j]);
124                 printf("\n");
125             }
126         }
127     }
128 }