HDU4347 The Closest M Points

n维空间最近m点的问题，应该算是经典模型了。

n维：

采用KD树解决，轮流沿着各个维度把点分成两部分（还有一种思路是把跨度最大的那一维度分成两部分，这一种方法是更优的，但是实现起来稍微复杂一点点。而均摊下来两种方法的复杂度相同，所以这里只采用了轮流的建树法），建树即可。

m点：

最近点和最近m点的问题是一样的。

#include <cstdio>
#include <cstring>
#include <cmath>
#include <iostream>
#include <vector>
#include <queue>
#include <stack>
#include <algorithm>
using namespace std;
typedef long long ll;

const int maxn = 50010;

int n, k, idx, q, m;

struct Point
{
    int x[5];
    bool operator < (const Point &o) const {return x[idx] < o.x[idx];}
}p[maxn], a, ans[15];

typedef pair<double, Point> tp;
priority_queue<tp> nq;

Point pt[maxn << 2];
int son[maxn << 2];

void buildKD(int l, int r, int rt = 1, int dep = 0)
{
    if (l > r) return;
    son[rt] = r - 1;
    son[rt<<1] = son[rt<<1|1] = -1;
    idx = dep % k;
    int mid = (l + r) >> 1;
    nth_element(p+l, p+mid, p+r+1);
    pt[rt] = p[mid];
    buildKD(l, mid-1, rt<<1, dep+1);
    buildKD(mid+1, r, rt<<1|1, dep+1);
}

void find(Point a, int m, int rt = 1, int dep = 0)
{
    if (son[rt] == -1) return;
    tp node(0, pt[rt]);
    for (int i = 0; i < k; i++) node.first += ((double)a.x[i]-pt[rt].x[i]) * ((double)a.x[i]-pt[rt].x[i]);
    int dim = dep % k, lhs = rt << 1, rhs = rt << 1 | 1, flag = 0;
    if (a.x[dim] >= pt[rt].x[dim]) swap(lhs, rhs);
    if (son[lhs] != -1) find(a, m, lhs, dep+1);
    if (nq.size() < m) nq.push(node), flag = 1;
    else
    {
        if (node.first < nq.top().first) nq.pop(), nq.push(node);
        if ((a.x[dim]-pt[rt].x[dim])*(a.x[dim]-pt[rt].x[dim]) < nq.top().first) flag = 1;
    }
    if (son[rhs] != -1 && flag) find(a, m, rhs, dep+1);
}

int main()
{
    while (scanf("%d%d", &n, &k) == 2)
    {
        for (int i=1;i<=n;i++)
            for (int j=0;j<k;j++)
                scanf("%d", &p[i].x[j]);
        buildKD(1, n);
        scanf("%d", &q);
        while (q--)
        {
            for (int i=0;i<k;i++) scanf("%d", &a.x[i]);
            scanf("%d", &m);
            while (!nq.empty()) nq.pop();
            find(a, m);
            for (int i=0;i<m;i++) ans[i] = nq.top().second, nq.pop();
            printf("the closest %d points are:\n", m);
            for (int i=m-1;i>=0;i--)
            {
                for (int j=0;j<k-1;j++) printf("%d ", ans[i].x[j]);
                printf("%d\n", ans[i].x[k-1]);
            }
        }
    }
    return 0;
}