"BZOJ3053" the Closest M Points kdtree good template One "Array Edition!!! 】

The multi-dimensional valuation is the same as the two dimensions, and the contribution of each dimension is identical to the two-dimensional ' x ', ' Y '.

Really do not understand how to think, there is a difference?!!

Details:

Mom eggs with multiple sets of data! I've been adjusting for a long time.

Then there is no space at the end of the line, otherwise the PE

To get rid of these, I am 1 A.

Unfortunately, there is no "removal" of this saying.

#include <cstdio> #include <cstring> #include <iostream> #include <algorithm> #define N 50100# Define K 10#define inf 0x3f3f3f3fusing namespace Std;int judge,n,m,q;struct point {int c[k];bool operator < (const poin T &a) Const{return C[judge]<a.c[judge];} int d (point &a) {int ret=0;for (int i=0;i<m;i++) ret+= (C[i]-a.c[i]) * (C[i]-a.c[i]); return ret;}} P[n];struct Ans{int x; Point P;bool operator < (const ANS &a) const {return x<a.x;}} Ans[15];struct Kdt{int son[n][2],cnt; Point C[n][2];void init () {Cnt=0;memset (son,0,sizeof (son));} int NewNode (point &p) {Cnt++,c[cnt][0]=c[cnt][1]=p;return cnt;} void update (int f) {int i,s;if (son[f][0]) {s=son[f][0];for (i=0;i<m;i++) c[f][0].c[i]=min (c[f][0].c[i],c[s][0].c[i ]), C[f][1].c[i]=max (C[f][1].c[i],c[s][1].c[i]);} if (Son[f][1]) {s=son[f][1];for (i=0;i<m;i++) c[f][0].c[i]=min (C[f][0].c[i],c[s][0].c[i]), C[f][1].c[i]=max (C[f] [1].c[i],c[s][1].c[i]);}} int Mindis (int f,point &p) {int ret=0,i;for (i=0;i<m;i++) {if (p.c[I]<c[f][0].c[i]) ret+= (C[f][0].c[i]-p.c[i]) * (C[f][0].c[i]-p.c[i]), else if (C[f][1].c[i]<p.c[i]) ret+= (c[f][ 1].c[i]-p.c[i]) * (C[f][1].c[i]-p.c[i]);} return ret;} int build (int l,int r,int jd=0) {int mid=l+r>>1,i;judge=jd;jd=jd+1>=m?jd-m+1:jd+1;nth_element (P+l,P+mid,P+r +1) int t=0;if (L&LT;MID) t=build (L,MID-1,JD); int X=newnode (P[mid]); Son[x][0]=t;if (Mid<r) son[x][1]=build (mid+1, R,JD); update (x); return x;} void query (int f,point &p,int rank) {int dis[3],i;dis[2]=p[f].d (p); if (ans[rank-1].x>dis[2]) ans[rank-1].x=dis[ 2],ans[rank-1].p=p[f],sort (Ans,ans+rank);D is[0]=dis[1]=0;if (son[f][0]) Dis[0]=mindis (son[f][0],p); if (son[f][1]) Dis[1]=mindis (son[f][1],p); int t=dis[0]>dis[1];if (son[f][t]&&dis[t]<ans[rank-1].x) query (Son[f][t] , P,rank); t^=1;if (son[f][t]&&dis[t]<ans[rank-1].x) query (Son[f][t],p,rank);}} Kdt;int Main () {//freopen ("test.in", "R", stdin); int i,j,k,root; Point X;while (~scanf ("%d%d", &n,&m)) {root=0;kdt.init (); for (i=1;i<=n;i++) for (j=0;j<m;j++) sCANF ("%d", &p[i].c[j]), if (n) root=kdt.build (1,n), for (scanf ("%d", &q), q--;) {for (i=0;i<m;i++) scanf ("%d", &x.c[i]); scanf ("%d", &k), for (i=0;i<15;i++) ans[i].x=inf;kdt.query (root,x,k);p rintf ("the closest%d Points are:\n ", k); for (i=0;i<k;i++) {for (j=0;j<m-1;j++) printf ("%d ", Ans[i].p.c[j]);p rintf ("%d\n ", ans[i].p.c[ M-1]);}} Puts ("");} return 0;}

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"BZOJ3053" the Closest M Points kdtree good template One "Array Edition!!! 】