[HDU 2966]In case of failure

Posted 2020-10-15 NaVi_Awson

Description

To help their clients deal with faulty Cash Machines, the board of The Planar Bank has decided to stick a label expressing sincere regret and sorrow of the bank about the failure on every ATM. The very same label would gently ask the customer to calmly head to the nearest Machine (that should hopefully
work fine).
In order to do so, a list of two-dimensional locations of all n ATMs has been prepared, and your task is to find for each of them the one closest with respect to the Euclidean distance.

Input

The input contains several test cases. The very first line contains the number of cases t (t <= 15) that follow. Each test cases begin with the number of Cash Machines n (2 <= n <= 10^5). Each of the next n lines contain the coordinates of one Cash Machine x,y (0 <= x,y <=10^9) separated by a space. No two
points in one test case will coincide.

Output

For each test case output n lines. i-th of them should contain the squared distance between the i-th ATM from the input and its nearest neighbour.

Sample Input

2 10 17 41 0 34 24 19 8 28 14 12 45 5 27 31 41 11 42 45 36 27 15 0 0 1 2 2 3 3 2 4 0 8 4 7 4 6 3 6 1 8 0 11 0 12 2 13 1 14 2 15 0

Sample Output

200 100 149 100 149 52 97 52 360 97 5 2 2 2 5 1 1 2 4 5 5 2 2 2 5

题目大意

求平面最近点对。

题解

裸的$KD-tree$。

我们用线性结构来存储这颗树，其主要思想是将一段区间内的元素中间那个元素作为根节点，其余作为左右子树，用个决策函数来划分。

询问的时候就比较暴力...其主要就是通过根节点以及其与子树的关系来优化搜索过程。

 1 //It is made by Awson on 2017.11.25
 2 #include <map>
 3 #include <set>
 4 #include <cmath>
 5 #include <ctime>
 6 #include <queue>
 7 #include <stack>
 8 #include <cstdio>
 9 #include <string>
10 #include <vector>
11 #include <cstdlib>
12 #include <cstring>
13 #include <iostream>
14 #include <algorithm>
15 #define LL long long
16 #define LD long double
17 #define Max(a, b) ((a) > (b) ? (a) : (b))
18 #define Min(a, b) ((a) < (b) ? (a) : (b))
19 #define dist(a, b, c, d) (sqr(a-c)+sqr(b-d))
20 #define sqr(x) ((x)*(x))
21 #define y1 yy
22 #define count COUNT
23 using namespace std;
24 const int N = 1e5;
25 
26 int n, now;
27 LL ans;
28 struct tt {
29   LL x[2];
30   bool operator < (const tt &b) const {
31     return x[now] < b.x[now];
32   }
33 }a[N+5], b[N+5];
34 struct KD_tree {
35   void build(int l, int r, int flag) {
36     if (l >= r) return;
37     now = flag;
38     int mid = (l+r)>>1;
39     nth_element(a+l, a+mid, a+r+1);
40     build(l, mid-1, flag^1); build(mid+1, r, flag^1);
41   }
42   void query(int l, int r, int flag, tt p) {
43     if (l > r) return;
44     int mid = (l+r)>>1;
45     LL tmp = dist(a[mid].x[0], a[mid].x[1], p.x[0], p.x[1]);
46     if (tmp && tmp < ans) ans = tmp;
47     if (a[mid].x[flag] < p.x[flag]) {
48       query(mid+1, r, flag^1, p);
49       if (ans > sqr(a[mid].x[flag]-p.x[flag])) 
50     query(l, mid-1, flag^1, p);
51     }else {
52       query(l, mid-1, flag^1, p);
53       if (ans > sqr(a[mid].x[flag]-p.x[flag])) 
54     query(mid+1, r, flag^1, p);
55     }
56   }
57 }kd;
58 
59 void work() {
60   scanf("%d", &n);
61   for (int i = 1; i <= n; i++) {
62     scanf("%lld%lld", &a[i].x[0], &a[i].x[1]);
63     b[i] = a[i];
64   }
65   kd.build(1, n, 0);
66   for (int i = 1; i <= n; i++) {
67     ans = 1e18;
68     kd.query(1, n, 0, b[i]);
69     printf("%lld\n", ans);
70   }
71 }
72 int main() {
73   int t; cin >> t;
74   while (t--) work();
75   return 0;
76 }