bzoj3053

 

k维最近点对。

要求输出前m近点对，用priority_queue存前几名就好，注意估价函数剪枝

 

#include<cstdio>#include<cmath>#include<cstring>#include<cstdlib>#include<algorithm>#include<queue>#define fi first#define se second#define pii pair<int,int>#define MK(a,b) make_pair((a),(b))#define sqr(x) (x)*(x) using namespace std;const int N=50005;inline int read(){int ans,f=1;char ch;while ((ch=getchar())<'0'||ch>'9') if (ch=='-') f=-1;ans=ch-'0';while ((ch=getchar())>='0'&&ch<='9') ans=ans*10+ch-'0';return ans*f;}int rt,n,k,D,m;struct aa{int l,r;int d[6],mi[6],mx[6];}a[N];void up(int u,int l){for (int i=0;i<k;i++) a[u].mi[i]=min(a[u].mi[i],a[l].mi[i]),a[u].mx[i]=max(a[u].mx[i],a[l].mx[i]);}bool cmp(aa a,aa b) {return a.d[D]<b.d[D];}int build(int l,int r,int dd){D=dd;int mid=(l+r)>>1;nth_element(a+l,a+mid,a+r+1,cmp);for (int i=0;i<k;i++)a[mid].mi[i]=a[mid].mx[i]=a[mid].d[i];if (l!=mid) a[mid].l=build(l,mid-1,(dd+1)%k),up(mid,a[mid].l);else a[mid].l=0;if (r!=mid) a[mid].r=build(mid+1,r,(dd+1)%k),up(mid,a[mid].r);else a[mid].r=0;return mid;}int x[6];priority_queue<pii> ans;int fdis(int u){int ans=0;for (int i=0;i<k;i++) ans+=sqr(x[i]-a[u].d[i]);return ans;}int f(int u){if (u==0) return 1e9;int ans=0;for (int i=0;i<k;i++){if (a[u].mi[i]>x[i]) ans+=sqr(a[u].mi[i]-x[i]);if (a[u].mx[i]<x[i]) ans+=sqr(x[i]-a[u].mx[i]);}return ans;}void query(int u){int dis=fdis(u);int dl=f(a[u].l),dr=f(a[u].r);if (ans.size()<m) ans.push(MK(dis,u));else if (ans.top().fi>dis) {ans.pop();ans.push(MK(dis,u));}if (dl<dr){if (ans.size()<m||ans.top().fi>dl) query(a[u].l);if (ans.size()<m||ans.top().fi>dr) query(a[u].r);}else {if (ans.size()<m||ans.top().fi>dr) query(a[u].r);if (ans.size()<m||ans.top().fi>dl) query(a[u].l);}}int tmp[15],cnt;void work(){for (int i=1;i<=n;i++)for (int j=0;j<k;j++) a[i].d[j]=read();rt=build(1,n,0);int t=read();while (t--){while (!ans.empty()) ans.pop();for (int i=0;i<k;i++) x[i]=read();m=read();query(rt);printf("the closest %d points are:\n",m);cnt=0;while (!ans.empty()) tmp[++cnt]=ans.top().se,ans.pop();for (int i=cnt;i>=1;i--){for (int j=0;j<k-1;j++) printf("%d ",a[tmp[i]].d[j]);printf("%d\n",a[tmp[i]].d[k-1]);}}}int main(){while (scanf("%d%d",&n,&k)!=EOF) work();return 0;}