【堆】【kd-tree】bzoj2626 JZPFAR
用堆记录答案。看看当前点是否比堆顶更优。
#include<cstdio> #include<queue> #include<cstring> #include<cmath> #include<algorithm> using namespace std; typedef double db; #define N 100001 #define EPS 0.0000001 #define INF 999999999999999999.0 #define KD 2//ά¶ÈÊý int qp[KD]; int n,root,kd=2,K; int dn; struct Ans { int p[KD],id; db d; Ans(){} Ans(int _p[],int _id,db _d){memcpy(p,_p,sizeof(p)); id=_id; d=_d;} }; bool operator < (const Ans &a,const Ans &b) {return fabs(a.d-b.d)>=EPS ? a.d>b.d : a.id<b.id;} priority_queue<Ans>Heap; db sqr(const int &x){return (db)x*(db)x;} struct Node { int minn[KD],maxx[KD],p[KD],id; 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+=sqr(max(0,qp[i]-minn[i])); t+=sqr(max(0,maxx[i]-qp[i])); } return sqrt(t); } }T[N]; 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+=sqr(a[i]-b[i]); return sqrt(t); } bool operator < (const Node &a,const Node &b){return a.p[dn]<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); if(Heap.size()<K) Heap.push(Ans(T[rt].p,T[rt].id,t)); else if(Heap.top().d-t<-EPS || (fabs(Heap.top().d-t)<EPS && T[rt].id<Heap.top().id)) { Heap.pop(); Heap.push(Ans(T[rt].p,T[rt].id,t)); } 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]-Heap.top().d>EPS || Heap.size()<K) && T[rt].ch[!f]) Query(T[rt].ch[!f]); if((dd[f]-Heap.top().d>EPS || Heap.size()<K) && T[rt].ch[f]) Query(T[rt].ch[f]); } int q; int main() { // freopen("bzoj2626.in","r",stdin); // freopen("bzoj3053.out","w",stdout); scanf("%d",&n); for(int i=1;i<=n;++i) { for(int j=0;j<kd;++j) scanf("%d",&T[i].p[j]); T[i].id=i; } Buildtree(); root=(1+n>>1); scanf("%d",&q); for(;q;--q) { while(!Heap.empty()) Heap.pop(); for(int i=0;i<kd;++i) scanf("%d",&qp[i]); scanf("%d",&K); Query(); printf("%d\n",Heap.top().id); } return 0; }
——The Solution By AutSky_JadeK From UESTC
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