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.
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.
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
Source
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 ",ans); 96 } 97 } 98 }