hdu2966 In case of failure（这道题就和ta的name一样，failure）

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
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

分析：
KDtree

tip

下方高能
我写好了之后，交上去就是WA

于是就从网上扒了一个AC代码对拍

狂拍不止，结果如下

疯了吗，没错就是WA，我放弃了。。。

这里写代码片
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define ll long long

using namespace std;

const ll INF=0x33333333;
struct node{
    int l,r,d[2],mn[2],mx[2],id;
};
node t[100010];
int root,n,x,y,cmpd,po[100010][2];
ll ans;

int cmp(const node &a,const node &b)
{
    return ((a.d[cmpd]<b.d[cmpd])||((a.d[cmpd]==b.d[cmpd])&&(a.d[!cmpd]<b.d[!cmpd])));
}

void update(int bh)
{
    int lc=t[bh].l;
    int rc=t[bh].r;
    if (lc)
    {
        t[bh].mn[0]=min(t[bh].mn[0],t[lc].mn[0]);
        t[bh].mn[1]=min(t[bh].mn[1],t[lc].mn[1]);
        t[bh].mx[0]=max(t[bh].mx[0],t[lc].mx[0]);
        t[bh].mx[1]=max(t[bh].mx[1],t[lc].mx[1]);
    }
    if (rc)
    {
        t[bh].mn[0]=min(t[bh].mn[0],t[rc].mn[0]);
        t[bh].mn[1]=min(t[bh].mn[1],t[rc].mn[1]);
        t[bh].mx[0]=max(t[bh].mx[0],t[rc].mx[0]);
        t[bh].mx[1]=max(t[bh].mx[1],t[rc].mx[1]);
    }
}

int build(int l,int r,int D)
{
    cmpd=D;
    int mid=(l+r)>>1;
    nth_element(t+l+1,t+mid+1,t+r+1,cmp);
    t[mid].mn[0]=t[mid].mx[0]=t[mid].d[0];
    t[mid].mn[1]=t[mid].mx[1]=t[mid].d[1];
    if (l!=mid) t[mid].l=build(l,mid-1,!D);
    if (r!=mid) t[mid].r=build(mid+1,r,!D);
    update(mid);
    return mid;
}

ll sqr(ll x) {return x*x;}
ll minn(ll a,ll b){if (a<b) return a;else return b;}

ll dis(int p,int x,int y)
{
    ll d=0;
    if (x<t[p].mn[0]) d+=sqr((ll)t[p].mn[0]-x);
    if (x>t[p].mx[0]) d+=sqr((ll)x-t[p].mx[0]);
    if (y<t[p].mn[1]) d+=sqr((ll)t[p].mn[1]-y);
    if (y>t[p].mx[1]) d+=sqr((ll)y-t[p].mx[1]);
    return d;
}

void ask(int now)
{
    ll d0,dl,dr;
    d0=sqr((ll)t[now].d[0]-x)+sqr((ll)t[now].d[1]-y);
    if (t[now].l) dl=dis(t[now].l,x,y);
    else dl=INF;
    if (t[now].r) dr=dis(t[now].r,x,y);
    else dr=INF;
    if (d0) ans=minn(ans,d0);
    if (dl<dr)
    {
        if (dl<ans) ask(t[now].l);
        if (dr<ans) ask(t[now].r);
    }
    else
    {
        if (dr<ans) ask(t[now].r);
        if (dl<ans) ask(t[now].l);
    }
}

ll an[150000];

int main()
{
    int T;
    scanf("%d",&T);
    while (T--)
    {
        memset(t,0,sizeof(t));
        scanf("%d",&n);
        for (int i=1;i<=n;i++)
            scanf("%d%d",&t[i].d[0],&t[i].d[1]),t[i].id=i;
        root=build(1,n,0);
        for (int i=1;i<=n;i++) 
        {
            x=t[i].d[0]; y=t[i].d[1];
            ans=INF;
            ask(root);
            an[t[i].id]=ans;
        }
        for (int i=1;i<=n;i++) printf("%lld
",an[i]); 
    }
    return 0;
}