[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);
}

推荐

http://www.cnblogs.com/xdruid/archive/2012/05/27/CP.html