HDU 4347 The Closest M Points

ゝ一世哀愁。 2022-05-14 06:27 134阅读 0赞

## 题目：[点击打开链接][Link 1] ##

## 题意：求取一个点的m个最近的点 ##

## 分析：kd树模板题。入门可参考[https://blog.csdn.net/u013534123/article/details/80952174][https_blog.csdn.net_u013534123_article_details_80952174]，代码细节可参考[https://blog.csdn.net/HackerTom/article/details/78198767][https_blog.csdn.net_HackerTom_article_details_78198767] ##

## 代码： ##

#pragma GCC optimize(2)
    #pragma GCC optimize(3)
    #pragma GCC optimize(4)
    #pragma comment(linker, "/STACK:102400000,102400000")
    #include<unordered_map>
    #include<unordered_set>
    #include<algorithm>
    #include<iostream>
    #include<fstream>
    #include<complex>
    #include<cstdlib>
    #include<cstring>
    #include<cassert>
    #include<iomanip>
    #include<string>
    #include<cstdio>
    #include<bitset>
    #include<vector>
    #include<cctype>
    #include<cmath>
    #include<ctime>
    #include<stack>
    #include<queue>
    #include<deque>
    #include<list>
    #include<set>
    #include<map>
    using namespace std;
    #define pt(a) cout<<a<<endl
    #define debug test
    #define mst(ss,b) memset((ss),(b),sizeof(ss))
    #define rep(i,a,n) for (int i=a;i<=n;i++)
    #define per(i,a,n) for (int i=n-1;i>=a;i--)
    #define all(x) (x).begin(),(x).end()
    #define fi first
    #define se second
    #define SZ(x) ((int)(x).size())
    #define ll long long
    #define ull unsigned long long
    #define pb push_back
    #define mp make_pair
    #define inf 0x3f3f3f3f
    #define eps 1e-10
    #define PI acos(-1.0)
    typedef pair<int,int> PII;
    const ll mod = 1e9+7;
    const int N = 1e5+10;
    
    ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
    ll qp(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
    int to[4][2]={
       {-1,0},{1,0},{0,-1},{0,1}};
    
    int n,m,k,t,q,idx,ex[N<<2];
    struct nd {
        int v[5];
        bool operator < (const nd &b) const {
            return v[idx]<b.v[idx];
        }
    }kdt[N<<2],p[N],aim;
    
    void build(int l,int r,int rt,int dep) {
        ex[rt]=1;
        ex[rt<<1]=ex[rt<<1|1]=0;
        idx=dep%k;
        int m=(l+r)>>1;
        nth_element(p+l,p+m,p+r+1);
        kdt[rt]=p[m];
        if(l<m) build(l,m-1,rt<<1,dep+1);
        if(r>m) build(m+1,r,rt<<1|1,dep+1);
    }
    
    typedef pair<int,nd> P;
    priority_queue<P> pq;
    
    void qy(int rt,int m,int dep) {
        if(!ex[rt]) return ;
        P now(0,kdt[rt]);
        rep(i,0,k-1) now.fi+=(aim.v[i]-kdt[rt].v[i])*(aim.v[i]-kdt[rt].v[i]);
        int d=dep%k;
        int lc=rt<<1,rc=rt<<1|1;
        if(aim.v[d]>=kdt[rt].v[d]) swap(lc,rc);
        if(ex[lc]) qy(lc,m,dep+1);
        int fd=0;
        if(pq.size()<m) {
            pq.push(now);
            fd=1;
        }else {
            if(now.fi<pq.top().fi) {
                pq.pop();
                pq.push(now);
            }
            if( (aim.v[d]-kdt[rt].v[d])*(aim.v[d]-kdt[rt].v[d]) < pq.top().fi ) fd=1;
        }
        if( ex[rc] && fd ) qy(rc,m,dep+1);
    }
    
    int main() {
        ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
        while(cin>>n>>k) {
            rep(i,0,n-1) rep(j,0,k-1) cin>>p[i].v[j];
            build(0,n-1,1,0);
            cin>>q;
            rep(i,1,q) {
                rep(j,0,k-1) cin>>aim.v[j];
                cin>>m;
                qy(1,m,0);
                int kk=0;
                while(!pq.empty()) p[kk++]=pq.top().se,pq.pop();
                cout<<"the closest "<<m<<" points are:"<<endl;
                for(int x=kk-1;x>=0;x--) {
                    for(int y=0;y<k-1;y++) cout<<p[x].v[y]<<" ";
                    cout<<p[x].v[k-1]<<endl;
                }
            }
        }
        return 0;
    }

[Link 1]: http://acm.hdu.edu.cn/showproblem.php?pid=4347
[https_blog.csdn.net_u013534123_article_details_80952174]: https://blog.csdn.net/u013534123/article/details/80952174
[https_blog.csdn.net_HackerTom_article_details_78198767]: https://blog.csdn.net/HackerTom/article/details/78198767