In case of failure

近点查询算法解析

最新推荐文章于 2025-09-07 20:32:33 发布

转载最新推荐文章于 2025-09-07 20:32:33 发布 · 204 阅读

0 ·

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/Fighting-sh/p/9873378.html

文章标签：

#java #php

本文深入探讨了在二维平面上寻找最近邻居点的算法，通过构建数据结构来快速定位每个现金机器的最近工作正常点，解决银行ATM故障时客户引导问题。采用优先队列和KD树进行高效查询，讲解了算法实现细节。

In case of failure

http://acm.hdu.edu.cn/showproblem.php?pid=2966

Time Limit: 60000/30000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2843 Accepted Submission(s): 1209

Problem 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

17 41

0 34

24 19

8 28

14 12

45 5

27 31

41 11

42 45

36 27

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

360

Source

Central European Programming Contest 2008

注意，最近点不能是自己，所以需要判cur.first!=0时才加入队列中

 1 #include<iostream>
 2 #include<queue>
 3 #include<cstring>
 4 #include<algorithm>
 5 #include<cstdio>
 6 #define N 100005
 7 using namespace std;
 8 
 9 int n,m,id;//n是点数,m是维度,id是当前切的维度
10 
11 struct sair{
12     long long p[5];
13     bool operator<(const sair &b)const{
14         return p[id]<b.p[id];
15     }
16 }_data[N],data[N<<3],tt[N];
17 int flag[N<<3];
18 
19 priority_queue<pair<long long,sair> >Q;
20 
21 void build(int l,int r,int rt,int dep){
22     if(l>r) return;
23     flag[rt]=1;
24     flag[rt<<1]=flag[rt<<1|1]=-1;
25     id=dep%m;
26     int mid=l+r>>1;
27     nth_element(_data+l,_data+mid,_data+r+1);
28     data[rt]=_data[mid];
29     build(l,mid-1,rt<<1,dep+1);
30     build(mid+1,r,rt<<1|1,dep+1);
31 }
32 
33 void query(sair p,int k,int rt,int dep){
34     if(flag[rt]==-1) return;
35     pair<long long,sair> cur(0,data[rt]);//获得当前节点
36     for(int i=0;i<m;i++){//计算当前节点到P点的距离
37         cur.first+=(cur.second.p[i]-p.p[i])*(cur.second.p[i]-p.p[i]);
38     }
39     int idx=dep%m;
40     int fg=0;
41     int x=rt<<1;
42     int y=rt<<1|1;
43     if(p.p[idx]>=data[rt].p[idx]) swap(x,y);
44     if(~flag[x]) query(p,k,x,dep+1);
45     //开始回溯
46     if(Q.size()<k){
47         if(cur.first!=0){
48             Q.push(cur);
49         }
50         fg=1;
51     }
52     else{
53         if(cur.first<Q.top().first){
54             if(cur.first!=0){
55                 Q.pop();
56                 Q.push(cur);
57             }
58         }
59         if((p.p[idx]-data[rt].p[idx])*(p.p[idx]-data[rt].p[idx])<Q.top().first){
60             fg=1;
61         }
62     }
63     if(~flag[y]&&fg){
64         query(p,k,y,dep+1);
65     }
66 }
67 
68 long long ans;
69 
70 
71 int main(){
72     int T;
73     scanf("%d",&T);
74     while(T--){
75         m=2;
76         scanf("%d",&n);
77         for(int i=1;i<=n;i++){
78             for(int j=0;j<m;j++){
79                 scanf("%lld",&_data[i].p[j]);
80                 tt[i].p[j]=_data[i].p[j];
81             }
82         }
83         build(1,n,1,0);
84         int k=1;
85         for(int i=1;i<=n;i++){
86             sair tmp;
87             for(int j=0;j<m;j++)
88                 tmp.p[j]=tt[i].p[j];
89             while(!Q.empty()){
90                 Q.pop();
91             }
92             query(tmp,k,1,0);
93             ans=Q.top().first;
94             Q.pop();
95             printf("%lld\n",ans);
96         }
97     }
98 }