[模板]平面最近点对
实现
将平面内点按$x$坐标排序,分治$x$坐标,设$ret=min(f(l,mid),f(mid+1,r))$,
将$x\in[mid-ret,mid+ret]$内的点按$y$坐标排序,算每个点与相邻的$6$个点的距离找最优解即可.
时间复杂度:$O(nlogn)$.
#define N 100005 #define INF 1e15 struct point{ double x,y; }p[N]; inline double sqr(double k){ return k*k; } inline double dis(point x,point y){ return sqrt(sqr(x.x-y.x)+sqr(x.y-y.y)); } inline bool cmpx(point x,point y){ if(x.x!=y.x) return x.x<y.x; return x.y<y.y; } inline bool cmpy(point x,point y){ if(x.y!=y.y) return x.y<y.y; return x.x<y.x; } inline double min_d(int l,int r){ double ret=INF; if(r-l<=20){ for(int i=l;i<r;++i) for(int j=i+1;j<=r;++j) ret=min(ret,dis(p[i],p[j])); return ret; } int mid=l+r>>1; ret=min(min_d(l,mid),min_d(mid+1,r)); while(p[l].x+ret<p[mid].x) ++l; while(p[r].x-ret>p[mid].x) --r; sort(p+l,p+1+r,cmpy); for(int i=l;i<r;++i) for(int j=min(r,i+6);j>i;--j) ret=min(ret,dis(p[i],p[j])); sort(p+l,p+1+r,cmpx); return ret; } inline double min_dis(){ sort(p+1,p+1+n,cmpx); return min_d(1,n); }
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