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;}