HDU #2966 In case of failure

Overview

给出平面上两两不重合的$n$个整点, 求每个点到它在其他$n-1$个点的最近临点的欧几里得距离的平方.

Solution

k-d tree 模板题.
关于k-d tree, 见这篇博客.

Implementation

#include <bits/stdc++.h>
#define lson id<<1
#define rson id<<1|1
#define sqr(x) (x)*(x)
using namespace std;
using LL=long long;
const int N=1e5+5;

// K-D tree: a special case of binary space partitioning trees
int DIM=2, idx;

struct Node{
    LL key[2];
    bool operator<(const Node &rhs)const{
        return key[idx]<rhs.key[idx];
    }
    void read(){
        for(int i=0; i<DIM; i++)
            scanf("%lld", key+i);
    }
    LL dis2(const Node &rhs)const{
        LL res=0;
        for(int i=0; i<DIM; i++)
            res+=sqr(key[i]-rhs.key[i]);
        return res;
    }
}p[N], _p[N];

Node a[N<<2];   // K-D tree
bool f[N<<2];

// [l, r)
void build(int id, int l, int r, int dep){
    if(l==r) return;    // error-prone
    f[id]=true, f[lson]=f[rson]=false;
    // select axis based on depth so that axis cycles through all valid values
    idx=dep%DIM;
    int mid=l+r>>1;
    // sort point list and choose median as pivot element
    nth_element(p+l, p+mid, p+r);
    a[id]=p[mid];
    build(lson, l, mid, dep+1);
    build(rson, mid+1, r, dep+1);
}

LL mi;

void query(const Node &p, int id, int dep){
    int dim=dep%DIM;
    int x=lson, y=rson;
    // left: <, right >=
    if(p.key[dim]>=a[id].key[dim])
        swap(x, y);

    if(f[x]) query(p, x, dep+1);
    LL cur=p.dis2(a[id]);
    if(cur && cur<mi) mi=cur;

    if(f[y] && sqr(a[id].key[dim]-p.key[dim])<mi)
        query(p, y, dep+1);
}

int main(){
    int T;
    scanf("%d", &T);
    for(int n; T--; ){
        scanf("%d", &n);
        for(int i=0; i<n; i++){
            p[i].read();
            _p[i]=p[i]; //error-prone
        }
        build(1, 0, n, 0);
        for(int i=0; i<n; i++){
            mi=LLONG_MAX;
            query(_p[i], 1, 0);  //error-prone
            printf("%lld
", mi);
        }
    }
    return 0;
}