BZOJ 3053: The Closest M Points(K-D Tree)
BZOJ 3053: The Closest M Points(K-D Tree)
Time Limit: 10 Sec Memory Limit: 128 MB
Submit: 1235 Solved: 418
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
真正意义上的的K-D Tree
就是把二维扩展到了$k$维
这样只需要在建树的时候按照维度循环建就可以了
#include<cstdio>
#include<queue>
#include<algorithm>
using namespace std;
const int MAXN = 1e6 + 10, INF = 1e9 + 10;
inline int read() {
char c = getchar(); int x = 0, f = 1;
while(c < '0' || c > '9') {if(c == '-')f = -1; c = getchar();}
while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
return x * f;
}
int N, K, WD, root;
int out[MAXN];
struct Point {
int x[6];
bool operator < (const Point &rhs) const {
return x[WD] < rhs.x[WD];
}
}P[MAXN], ask;
#define ls(x) T[x].ls
#define rs(x) T[x].rs
struct KDTree {
int mn[6], mx[6], ls, rs;
Point tp;
}T[MAXN];
struct Ans {
int val, ID;
bool operator < (const Ans &rhs) const{
return val < rhs.val;
}
};
priority_queue<Ans>Q;
int sqr(int x) {
return x * x;
}
void update(int k) {
for(int i = 1; i <= K; i++) {
T[k].mn[i] = T[k].mx[i] = T[k].tp.x[i];
if(ls(k)) T[k].mn[i] = min(T[k].mn[i], T[ls(k)].mn[i]), T[k].mx[i] = max(T[k].mx[i], T[ls(k)].mx[i]);
if(rs(k)) T[k].mn[i] = min(T[k].mn[i], T[rs(k)].mn[i]), T[k].mx[i] = max(T[k].mx[i], T[rs(k)].mx[i]);
}
}
int Build(int l, int r, int wd) {
WD = wd;
if(l > r) return 0;
int mid = l + r >> 1;
nth_element(P + l, P + mid, P + r + 1);
T[mid].tp = P[mid];
T[mid].ls = Build(l, mid - 1, (wd + 1) % K);
T[mid].rs = Build(mid + 1, r, (wd + 1) % K);
update(mid);
return mid;
}
int GetMinDis(Point a, KDTree b) {
//if(b) return INF;
int ans = 0;
for(int i = 1; i <= K; i++) {
if(a.x[i] < b.mn[i]) ans += sqr(b.mn[i] - a.x[i]);
if(a.x[i] > b.mx[i]) ans += sqr(a.x[i] - b.mx[i]);
}
return ans;
}
int Dis(Point a, Point b) {
int ans = 0;
for(int i = 1; i <= K; i++)
ans += sqr(abs(a.x[i] - b.x[i]));
return ans;
}
void Query(int k) {
int ans = Dis(ask, T[k].tp);
if(ans < Q.top().val) Q.pop(), Q.push((Ans){ans, k});
int disl = INF, disr = INF;
if(ls(k)) disl = GetMinDis(ask, T[ls(k)]);
if(rs(k)) disr = GetMinDis(ask, T[rs(k)]);
if(disl < disr) {
if(disl < Q.top().val) Query(ls(k));
if(disr < Q.top().val) Query(rs(k));
}
else {
if(disr < Q.top().val) Query(rs(k));
if(disl < Q.top().val) Query(ls(k));
}
}
main() {
while(scanf("%d %d", &N, &K) != EOF) {
for(int i = 1; i <= N; i++)
for(int j = 1; j <= K; j++)
P[i].x[j] = read();
root = Build(1, N, 0);
int T = read();
while(T--) {
for(int i = 1; i <= K; i++) ask.x[i] = read();
int M = read();
printf("the closest %d points are:\n", M);
for(int i = 1; i <= M; i++) Q.push((Ans){INF, 0});
Query(root);
for(int i = 1; i <= M; i++)
out[i] = Q.top().ID, Q.pop();
for(int i = M; i >= 1; i--)
for(int j = 1; j <= K; j++)
printf("%d%c", P[out[i]].x[j], j != K ? ' ' : '\n');
}
}
}