BZOJ 3053: The Closest M Points(K-D Tree)

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?

軟工學院的課程很討厭！ZLC同志遇到了一個頭疼的問題：在K維空間裡面有許多的點，對於某些給定的點，ZLC需要找到和它最近的m個點。

（這裡的距離指的是歐幾裡得距離：D(p, q) = D(q, p) =  sqrt((q1 - p1) ^ 2 + (q2 - p2) ^ 2 + (q3 - p3) ^ 2 + ... + (qn - pn) ^ 2)

ZLC要去打Dota，所以就麻煩你幫忙解決一下了……

【Input】

第一行，兩個非負整數：點數n(1 <= n <= 50000)，和維度數k(1 <= k <= 5)。
接下來的n行，每行k個整數，代表一個點的坐標。
接下來一個正整數：給定的詢問數量t(1 <= t <= 10000)
下麵2*t行：
　　第一行，k個整數：給定點的坐標
　　第二行：查詢最近的m個點(1 <= m <= 10)

所有坐標的絕對值不超過10000。
有多組數據！

【Output】

對於每個詢問，輸出m+1行：
第一行："the closest m points are:" m為查詢中的m
接下來m行每行代表一個點，按照從近到遠排序。

保證方案唯一，下麵這種情況不會出現：
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

Sample Output

HINT

Source

k-d tree

#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');          
        }
    }
}