"Kd-tree" bzoj3053 the Closest M Points

With p2626. Because k is small, you do not need to use heaps.

#include <cstdio> #include <cstring> #include <cmath> #include <algorithm>using namespace std;     typedef double DB; #define N 50001#define INF 2147483647.0#define KD 5//èêýint qp[kd];int n,root,kd,k;int dn;struct Ans{    int P[KD]; DB D;} Ans[10];int sqr (const int &x) {return x*x;}    struct node{int MINN[KD],MAXX[KD],P[KD];    int ch[2];      void Init () {for (int i=0;i<kd;++i) minn[i]=maxx[i]=p[i];        } db Dis () {db t=0;            for (int i=0;i<kd;++i) {t+= (db) Sqr (Max (0,minn[i]-qp[i]));          t+= (db) Sqr (Max (0,qp[i]-maxx[i]));      } return sqrt (t);          }}t[n<<1];void Update (int rt) {for (Int. i=0;i<2;++i) if (T[rt].ch[i]) for (int j=0;j<kd;++j)            {t[rt].minn[j]=min (t[rt].minn[j],t[t[rt].ch[i]].minn[j]);          T[rt].maxx[j]=max (T[rt].maxx[j],t[t[rt].ch[i]].maxx[j]);    }}DB Dis (int a[],int b[]) {db t=0;  for (int i=0;i<kd;++i)    t+= (db) Sqr (A[i]-b[i]); return sqrt (t);} BOOL operator < (const node &a,const node &b) {return A.P[DN]&LT;B.P[DN];}    int buildtree (int l=1,int r=n,int d=0) {dn=d;    int m= (l+r>>1);    Nth_element (t+l,t+m,t+r+1); T[M].    Init ();    if (l!=m) T[m].ch[0]=buildtree (L,m-1, (d+1)%KD);    if (m!=r) T[m].ch[1]=buildtree (M+1,r, (d+1)%KD);    Update (m); return m;}    void Query (int rt=root) {db T=dis (T[RT].P,QP);          for (int i=0;i<k;++i) if (T&LT;ANS[I].D) {for (int j=k-1;j>=i+1;--j) ans[j]=ans[j-1];          ans[i].d=t;          memcpy (ans[i].p,t[rt].p,sizeof (T[RT].P));        Break    } DB dd[2]; for (int i=0;i<2;i++) if (T[rt].ch[i]) dd[i]=t[t[rt].ch[i]].      Dis ();    else Dd[i]=inf;    BOOL F= (dd[0]<=dd[1]);    if (dd[!f]<ans[k-1].d && t[rt].ch[!f]) Query (t[rt].ch[!f]); if (dd[f]<ans[k-1].d && t[rt].ch[f]) Query (t[rt].ch[f]);} int Q;int Main () {//Freopen ("Bzoj3053.in", "R", stdin);//FreOpen ("Bzoj3053.out", "w", stdout);            while (scanf ("%d%d", &AMP;N,&AMP;KD)!=eof) {for (Int. i=1;i<=n;++i) for (int j=0;j<kd;++j)        scanf ("%d", &t[i].p[j]);        Buildtree ();        Root= (1+n>>1);        scanf ("%d", &q);            for (; q;--q) {for (int i=0;i<kd;++i) scanf ("%d", &qp[i]);            scanf ("%d", &k);            for (int i=0;i<k;++i) Ans[i].d=inf;            Query ();            printf ("The closest%d points are:\n", K); for (int i=0;i<k;++i) {for (int j=0;j<kd-1;++j) printf ("%d", Ans[i].p[j]                );              printf ("%d\n", ans[i].p[kd-1]);      }} for (int i=1;i<=n;++i) t[i].ch[0]=t[i].ch[1]=0; } return 0;}

"Kd-tree" bzoj3053 the Closest M Points