KDTree模板，在m维空间中找最近的k个点，用的是欧几里德距离。
理解了好久，昨晚始终不明白那些“估价函数”，后来才知道分情况讨论，≤k还是=k，在当前这一维度距离过线还是不过线，过线则要继续搜索另一个子树。还有别忘了当前这个节点！
#include<cmath>
#include<queue>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define read(x) x=getint()
using namespace std;
typedef long long LL;
const int N = 50003;
const int inf = 0x7fffffff;
int getint() {
int k = 0, fh = 1; char c = getchar();
for(; c < '0' || c > '9'; c = getchar())
if (c == '-') fh = -1;
for(; c >= '0' && c <= '9'; c = getchar())
k = k * 10 + c - '0';
return k * fh;
}
int n, m, root, D;
LL minn;
struct P {
int d[5], mx[5], mn[5], l, r, id;
P (): l(0), r(0), id(0) {};
int &operator [] (int x) {return d[x];}
bool operator < (P point) const {return d[D] < point[D];}
} T[N << 1], po[N];
priority_queue <pair <LL, int> > Q;
LL sqr(LL x) {return x * x;}
LL dis(P a, P b) {
LL ret = 0;
for(int i = 0; i < m; ++i) ret += sqr(a[i] - b[i]);
return ret;
}
void pushup(int x, int y) {
for(int i = 0; i < m; ++i)
T[x].mn[i] = min(T[x].mn[i], T[y].mn[i]),
T[x].mx[i] = max(T[x].mx[i], T[y].mx[i]);
}
int Build(int l, int r, int dd) {
D = dd; int mid = (l + r) >> 1; nth_element(po + l, po + mid, po + r + 1);
for(int i = 0; i < m; ++i)
T[mid].mn[i] = T[mid].mx[i] = T[mid].d[i] = po[mid].d[i];
T[mid].id = mid;
if (l < mid) T[mid].l = Build(l, mid - 1, (dd + 1) % m);
if (mid < r) T[mid].r = Build(mid + 1, r, (dd + 1) % m);
if (T[mid].l) pushup(mid, T[mid].l);
if (T[mid].r) pushup(mid, T[mid].r);
return mid;
}
void ask(int rt, int dd, P p, int k) {
int L = T[rt].l, R = T[rt].r;
if (p[dd] >= T[rt][dd]) swap(L, R);
if (L) ask(L, (dd + 1) % m, p, k);
bool pd = 0; LL di = dis(T[rt], p); minn = min(minn, di);
if (Q.size() < k) { Q.push(pair <LL, int> (di, rt)); pd = 1;}
else {
if (di < Q.top().first) Q.pop(), Q.push(make_pair(di, rt));
if (sqr(p[dd] - T[rt][dd]) < Q.top().first) pd = 1;
}
if (pd && R) ask(R, (dd + 1) % m, p, k);
}
int ans[N];
int main() {
while (~scanf("%d%d", &n, &m)) {
for(int i = 1; i <= n; ++i)
for(int j = 0; j < m; ++j)
read(po[i][j]);
memset(T, 0, sizeof(T));
root = Build(1, n, 0);
int qq; read(qq);
for(; qq; --qq) {
P p; int k;
for(int i = 0; i < m; ++i)
read(p[i]);
read(k);
printf("the closest %d points are:\n", k);
minn = inf; ask(root, 0, p, k);
while (!Q.empty()) {
ans[++ans[0]] = Q.top().second;
Q.pop();
}
for(; ans[0]; --ans[0])
for(int i = 0; i < m; ++i)
printf("%d%c", T[ans[ans[0]]][i]," \n"[i == m - 1]);
}
}
return 0;
}
我就是弱啊～～～