hdu 2966 In case of failure ( kd-tree(只有查询) 模板题)

题目链接:hdu2966  

题意:

给出二维平面上n(1E5)个点,问对于每个点,其他距离其最近的点的距离是多少。

思路:

kd-tree 裸题。

kd-tree 学习笔记

  1/* ***********************************************
  2Author :111qqz
  3Created Time :2017年10月08日 星期日 18时43分38秒
  4File Name :2996.cpp
  5************************************************ */
  6
  7#include <cstdio>
  8#include <cstring>
  9#include <iostream>
 10#include <algorithm>
 11#include <vector>
 12#include <queue>
 13#include <set>
 14#include <map>
 15#include <string>
 16#include <cmath>
 17#include <cstdlib>
 18#include <ctime>
 19#define PB push_back
 20#define fst first
 21#define sec second
 22#define lson l,m,rt<<1
 23#define rson m+1,r,rt<<1|1
 24#define ms(a,x) memset(a,x,sizeof(a))
 25typedef long long LL;
 26#define pi pair < int ,int >
 27#define MP make_pair
 28
 29using namespace std;
 30const double eps = 1E-8;
 31const int dx4[4]={1,0,0,-1};
 32const int dy4[4]={0,-1,1,0};
 33const int inf = 0x3f3f3f3f;
 34const int N=1E5+7;
 35int n;
 36struct Point
 37{
 38    LL x,y;
 39}p[N],p2[N]; //复制一份,因为nth_element的时候会把顺序打乱。
 40bool dv[N]; //划分方式
 41bool cmpx( const Point & p1, const Point &p2)
 42{
 43    return p1.x<p2.x;
 44}
 45bool cmpy(const Point &p1 ,const Point &p2)
 46{
 47    return p1.y<p2.y;
 48}
 49LL getDis( Point a, Point b)
 50{
 51    return (a.x-b.x)*(a.x-b.x) + (a.y-b.y)*(a.y-b.y);
 52}
 53void build ( int l,int r)
 54{
 55    if (l>r) return;
 56    int mid = (l+r) >> 1;
 57    int minx = min_element(p+l,p+r,cmpx)->x;
 58    int miny = min_element(p+l,p+r,cmpy)->y;
 59    int maxx = max_element(p+l,p+r,cmpx)->x;
 60    int maxy = max_element(p+l,p+r,cmpy)->y;
 61    dv[mid] = maxx-minx >= maxy-miny;
 62    nth_element(p+l,p+mid,p+r+1,dv[mid]?cmpx:cmpy);
 63    build(l,mid-1);
 64    build(mid+1,r);
 65
 66}
 67LL res;
 68void query( int l,int r,Point a)
 69{
 70    if (l>r) return;
 71    int mid = (l+r)>>1;
 72    LL dis = getDis(a,p[mid]);
 73    //printf("%lld %lld  %lld %lld    %lld\n",a.x,a.y,p[mid].x,p[mid].y,dis);
 74    if (dis>0) res = min(res,dis);//判掉和自己的距离。 
 75    LL d = dv[mid]?(a.x-p[mid].x):(a.y-p[mid].y);
 76    int l1=l,r1=mid-1,l2=mid+1,r2=r;
 77    if (d>0) swap(l1,l2),swap(r1,r2);
 78    query(l1,r1,a); //左儿子
 79    if (d*d<res) query(l2,r2,a);//如果与分界线相交,则也要查询右儿子。
 80}
 81
 82int main()
 83{
 84    #ifndef  ONLINE_JUDGE 
 85    freopen("./in.txt","r",stdin);
 86  #endif
 87    int T;
 88    cin>>T;
 89    while (T--)
 90    {
 91        scanf("%d",&n);
 92        for ( int i = 1 ; i <= n ; i++) {
 93        scanf("%lld %lld",&p[i].x,&p[i].y);
 94        p2[i] = p[i];
 95        }
 96        build (1,n);
 97        for ( int i = 1 ; i <= n ; i++)
 98        {
 99        res = 1LL<<60;
100        query(1,n,p2[i]);
101        printf("%lld\n",res);
102        }
103
104    }
105  #ifndef ONLINE_JUDGE  
106  fclose(stdin);
107  #endif
108    return 0;
109}