【K-D树 在限制条件下求最近欧几里德距离】HDU

Problem Description

给你n个酒店，m个人。
分别给你每个酒店的坐标，和酒店的价格。
分别给你每个人的坐标，和人最多能够承受的价格。
对于每个人，让你求这个人能够承受价格的酒店，那个距离他最近。

思路：

求最近距离的时候，加个条件限制即可。

#include<bits/stdc++.h>using namespace std;#define ll long longconst int DIM = 3;//三维，第三维是价格const int MAX = 200055;const ll inf = 0x3f3f3f3f3f3f3f3f;struct node{    int l, r;    int id;//用来记录输入顺序    int d[DIM], maxn[DIM], minn[DIM];    inline void maintain()//初始化    {        l = r = 0;        for(int i = 0; i < DIM; i++)            maxn[i] = minn[i] = d[i];    }}tree[MAX];int D;bool operator < (const node &x, const node &y)//重载从小到大{    return x.d[D] < y.d[D];}inline void Merge(int o)//归并，向上更新{    int son[2] = {tree[o].l, tree[o].r};    for(int i = 0; i < 2; i++)    {        if(!son[i]) continue;        for(int j = 0; j < DIM; j++)        {            tree[o].maxn[j] = max(tree[o].maxn[j], tree[son[i]].maxn[j]);            tree[o].minn[j] = min(tree[o].minn[j], tree[son[i]].minn[j]);        }    }}int build(int l, int r, int now)//建树{    int mid = (l+r)>>1;    D = now;    nth_element(tree+l, tree+mid, tree+r+1);    tree[mid].maintain();    if(l < mid) tree[mid].l = build(l, mid-1, (now+1)%DIM);    if(r > mid) tree[mid].r = build(mid+1, r, (now+1)%DIM);    Merge(mid);    return mid;}inline ll sqr(int x){    return (ll)x * (ll)x;}inline ll dis(int o, int k)//求两点间距离的平方{    return sqr(tree[o].d[0] - tree[k].d[0]) + sqr(tree[o].d[1] - tree[k].d[1]);}inline ll partionMin(int o, int k)//k->o点 可能的最近距离{    if(tree[o].minn[2] > tree[k].d[2]) return inf;//价格不满足，距离返回最远    ll red = 0;    for(int i = 0; i < DIM - 1; i++)    {        if(tree[k].d[i] > tree[o].maxn[i]) red += sqr(tree[o].maxn[i] - tree[k].d[i]);        if(tree[k].d[i] < tree[o].minn[i]) red += sqr(tree[o].minn[i] - tree[k].d[i]);    }    return red;}ll ans;int ans_id, ans_node;void query(int o, int k)//求满足条件最近距离的酒店{   // printf("%d %d %d\n", tree[o].d[0], tree[o].d[1], tree[o].d[2]);    ll dm = dis(o, k);//o k两点距离的平方    if(tree[o].d[2] <= tree[k].d[2] && (dm < ans || (dm == ans && tree[o].id < ans_id)))//满足条件，更新    {        ans_id = tree[o].id;        ans = dm;        ans_node = o;    }    ll dl = tree[o].l ? partionMin(tree[o].l, k) : inf;    ll dr = tree[o].r ? partionMin(tree[o].r, k) : inf;    if(dl < dr)//优化时间    {        if(dl <= ans) query(tree[o].l, k);        if(dr <= ans) query(tree[o].r, k);    }    else    {        if(dr <= ans) query(tree[o].r, k);        if(dl <= ans) query(tree[o].l, k);    }}int main(){    int T, n, m, i, j;    scanf("%d", &T);    while(T--)    {        scanf("%d %d", &n, &m);        for(i = 1; i <= n; i++){            for(j = 0; j < DIM; j++)            scanf("%d", &tree[i].d[j]);            tree[i].id = i;        }        int root = build(1, n, 0);        while(m--)        {            for(j = 0; j < DIM; j++)                scanf("%d", &tree[n+1].d[j]);            ans = inf;            query(root, n+1);          //  printf("%d\n", ansid);            for(j = 0; j < DIM; j++)            {                if(j) printf(" ");                printf("%d", tree[ans_node].d[j]);            }            printf("\n");        }    }    return 0;}