HDU4347-The Closest M Points

The Closest M Points

                                                              Time Limit: 16000/8000 MS (Java/Others)    Memory Limit: 98304/98304 K (Java/Others)
                                                                                        Total Submission(s): 5507    Accepted Submission(s): 1743

Problem 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 = (p ₁, p ₂,..., p _n) and q = (q ₁, q ₂,..., q _n) are two points in Euclidean n-space, then the distance from p to q, or from q to p is given by:

Can you help him solve this problem?

 

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
  

 

Author

HIT

 

Source

2012 Multi-University Training Contest 5

 

题意：给你n个k维的点，有q次询问，每次给你一个点，求出离它最近的m个点

解题思路：kd-tree+优先队列

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cctype>
#include <map>
#include <cmath>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <functional>

using namespace std;

#define LL long long
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int N = 50000 + 5;
const int demension = 5;//二维

struct node
{
	int pos[demension];
	int ma[demension], mi[demension];
	int l, r;
	LL dis;
	bool operator<(const node &a)const
	{
		return dis < a.dis;
	}
}a[N], x, pre, ans[15];
priority_queue<node>q;
int cmpDem;//以第cmpDem维作比较
int root, n, qq, K, m, Size;

bool cmp(const node &a, const node&b)
{
	return a.pos[cmpDem] < b.pos[cmpDem];
}

void Merge(int k)
{
	for (int i = 0; i < K; i++)
	{
		if (a[k].l)
		{
			a[k].ma[i] = max(a[k].ma[i], a[a[k].l].ma[i]);
			a[k].mi[i] = min(a[k].mi[i], a[a[k].l].mi[i]);
		}
		if (a[k].r)
		{
			a[k].ma[i] = max(a[k].ma[i], a[a[k].r].ma[i]);
			a[k].mi[i] = min(a[k].mi[i], a[a[k].r].mi[i]);
		}
	}
}

LL dis(int k)
{
	LL dis = 0;
	for (int i = 0; i < K; i++)
	{
		if (x.pos[i] < a[k].mi[i]) dis += 1LL * (a[k].mi[i] - x.pos[i])*(a[k].mi[i] - x.pos[i]);
		if (x.pos[i] > a[k].ma[i]) dis += 1LL * (x.pos[i] - a[k].ma[i])*(x.pos[i] - a[k].ma[i]);
	}
	return dis;
}

int build(int l, int r, int k)
{
	if (l > r) return 0;
	int mid = (l + r) / 2;
	//以第mid个元素为中心排序
	cmpDem = k;
	nth_element(a + l, a + mid, a + r + 1, cmp);
	//左右子树
	a[mid].l = build(l, mid - 1, (k + 1) % demension);
	a[mid].r = build(mid + 1, r, (k + 1) % demension);
	Merge(mid);
	return mid;
}

void query(int k)
{
	a[k].dis = 0;
	for (int i = 0; i < K; i++) a[k].dis += 1LL * (x.pos[i] - a[k].pos[i])*(x.pos[i] - a[k].pos[i]);
	if (Size < m) q.push(a[k]), Size++;
	else
	{
		pre = q.top();
		if (pre.dis > a[k].dis) q.pop(), q.push(a[k]);
	}
	LL dl = a[k].l ? dis(a[k].l) : INF, dr = a[k].r ? dis(a[k].r) : INF;
	if (min(dl, dr) >q.top().dis&&Size >= m) return;
	if (dl < dr)
	{
		if (Size < m || dl < q.top().dis) query(a[k].l);
		if (Size < m || dr < q.top().dis) query(a[k].r);
	}
	else
	{
		if (Size < m || dr < q.top().dis) query(a[k].r);
		if (Size < m || dl < q.top().dis) query(a[k].l);
	}
}

int main()
{
	while (~scanf("%d%d", &n, &K))
	{
		for (int i = 1; i <= n; i++)
		{
			for (int j = 0; j < K; j++) scanf("%d", &a[i].pos[j]), a[i].mi[j] = a[i].ma[j] = a[i].pos[j];
			a[i].l = a[i].r = 0;
		}
		root = build(1, n, 0);
		scanf("%d", &qq);
		while (qq--)
		{
			for (int i = 0; i < K; i++) scanf("%d", &x.pos[i]);
			scanf("%d", &m);
			Size = 0;
			query(root);
			printf("the closest %d points are:\n", m);
			for (int j = m; j >= 1; j--)
			{
				ans[j] = q.top();
				q.pop();
			}
			for (int i = 1; i <= m; i++)
			{
				for (int j = 0; j < K - 1; j++) printf("%d ", ans[i].pos[j]);
				printf("%d\n", ans[i].pos[K - 1]);
			}
		}
	}
	return 0;
}