The Closest M Points【K-D Tree+堆】

题目链接 HDU - 4347

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  给出N个点的K维图，然后有T次询问，每次询问一个点（x，y），问它的前M临近的点按顺序分别是哪几个，保证答案的唯一性。

  于是，根据这里的M很小，所以想到的就是拿一个堆来维护，值大的在前，我们尽可能使得堆的最大值最小，同时就是保证了堆内所有的元素的最小性质，所以一开始的时候push进去M个INF。

  剩下的，就是KD Tree的查询了，每次输入一个点，不用管是不是和已有点的重合问题。重合就当距离是0，也是可行的。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <bitset>
//#include <unordered_map>
//#include <unordered_set>
#define lowbit(x) ( x&(-x) )
#define pi 3.141592653589793
#define e 2.718281828459045
#define INF 0x3f3f3f3f3f3f3f3f
#define eps 1e-8
#define HalF (l + r)>>1
#define lsn rt<<1
#define rsn rt<<1|1
#define Lson lsn, l, mid
#define Rson rsn, mid+1, r
#define QL Lson, ql, qr
#define QR Rson, ql, qr
#define myself rt, l, r
#define MP(a, b) make_pair(a, b)
#define MAX_3(a, b, c) max(a, max(b, c))
#define Rabc(x) x > 0 ? x : -x
using namespace std;
typedef unsigned long long ull;
typedef unsigned int uit;
typedef long long ll;
const int maxK = 5, maxN = 5e4 + 7;
int N, K, Ques, M;
struct node
{
    ll d[maxK]; int id;
    node(ll a=0, ll b=0, int c=0):d{a, b}, id(c) {}
    friend bool operator < (node e1, node e2) { return e1.id < e2.id; }
    void In() { for(int i=0; i<K; i++) scanf("%lld", &d[i]); }
} a[maxN], b[maxN];
int op;
inline bool cmp(node e1, node e2) { return e1.d[op] < e2.d[op]; }
node q, point;
ll dis(node a, node b)
{
    ll sum = 0;
    for(int i=0; i<K; i++) sum += (a.d[i] - b.d[i]) * (a.d[i] - b.d[i]);
    return sum;
}
priority_queue<pair<ll, node>> Q;
struct KD_Close
{
    node tree[maxN << 2];
    int key[maxN << 2];
    double var[maxN << 2];
    void build(int rt, int l, int r)
    {
        if(l > r) return;
        op = 0; key[rt] = 0;
        for(int i=0; i<K; i++)
        {
            double ave = 0.;
            var[i] = 0;
            for(int j=l; j<=r; j++) ave += a[j].d[i];
            ave /= (r - l + 1.);
            for(int j=l; j<=r; j++) var[i] += (ave - a[j].d[i]) * (ave - a[j].d[i]);
            var[i] /= (r - l + 1.);
            if(var[i] > var[key[rt]])
            {
                key[rt] = i;
                op = i;
            }
        }
        int mid = HalF;
        nth_element(a + l, a + mid, a + r + 1, cmp);
        tree[rt] = a[mid];
        build(rt << 1, l, mid - 1); build(Rson);
    }
    void query(int rt, int l, int r)
    {
        if(l > r) return;
        int mid = HalF;
        ll dist = dis(q, tree[rt]);
        if(dist < Q.top().first)
        {
            Q.pop(); Q.push(MP(dist, tree[rt]));
            point = tree[rt];
        }
        int k_key = key[rt];
        ll ra = (q.d[k_key] - tree[rt].d[k_key]) * (q.d[k_key] - tree[rt].d[k_key]);
        if(q.d[k_key] < tree[rt].d[k_key])
        {
            query(rt << 1, l, mid - 1);
            if(ra < Q.top().first) query(Rson);
        }
        else
        {
            query(Rson);
            if(ra < Q.top().first) query(rt << 1, l, mid - 1);
        }
    }
} close_Tree;
node Stap[15];
int Stop;
int main()
{
    while(scanf("%d%d", &N, &K) != EOF)
    {
        for(int i=1; i<=N; i++)
        {
            a[i].In();
            a[i].id = i;
            b[i] = a[i];
        }
        close_Tree.build(1, 1, N);
        scanf("%d", &Ques);
        while(Ques--)
        {
            q.In();
            scanf("%d", &M);
            for(int i=1; i<=M; i++) Q.push(MP(INF, node()));
            close_Tree.query(1, 1, N);
            Stop = 0;
            while(!Q.empty())
            {
                Stap[++Stop] = Q.top().second;
                Q.pop();
            }
            printf("the closest %d points are:\n", M);
            for(int i=Stop; i; i--)
            {
                for(int j=0; j<K; j++) printf("%lld%c", Stap[i].d[j], (j == K - 1 ? '\n' : ' '));
            }
        }
    }
    return 0;
}