hdu 1007 Quoit Design(分治法求最近点对)

来源：互联网发布：陈江华12年奥运会数据编辑：程序博客网时间：2024/05/21 09:24

大致题意：给N个点，求最近点对的距离 d ；输出：r = d/2。

// Time 2093 ms; Memory 1812 K

#include<iostream>#include<cstdio>#include<cmath>#include<algorithm>#define eps 1e-8#define maxn 100010#define sqr(a) ((a)*(a))using namespace std;int sig(double x){return (x>eps)-(x<-eps);}struct point{double x,y;point(double xx=0,double yy=0):x(xx),y(yy){}}p[maxn];bool operator < (point a,point b){return a.x<b.x || (a.x==b.x && a.y<b.y);}bool cmp(point a,point b){return a.y<b.y || (a.y==b.y && a.x<b.x);}double len(point a,point b){return sqrt(sqr(a.x-b.x)+sqr(a.y-b.y));}double min_dis(int l,int r){int i,j,k,s;double d,mi=-1;if(r-l<4){for(i=l;i<r;i++) for(j=i+1;j<r;j++){d=len(p[i],p[j]);if(mi<0 || sig(d-mi)<0) mi=d;}return mi;}int mid=(l+r)/2;double minl=min_dis(l,mid);double minr=min_dis(mid,r);mi=minl<minr?minl:minr;for(i=1;i<mid-l && sig(p[mid].x-p[mid-i].x-mi)<0;i++);i--;for(j=1;j<r-mid && sig(p[mid+j].x-p[mid].x-mi)<0;j++);j--;sort(p+mid-i,p+mid,cmp);sort(p+mid,p+mid+j+1,cmp);int t=mid,flag;for(k=mid-i;k<mid;k++) {flag=1;for(s=t;s<=mid+j;s++){if(sig(p[k].y-p[s].y)>0){if(sig(p[k].y-p[s].y-mi)>=0) continue;else{if(flag){flag=0;t=s;}d=len(p[s],p[k]);if(sig(d-mi)<0) mi=d;}}else{if(sig(p[s].y-p[k].y-mi)>=0) break;else{d=len(p[s],p[k]);if(sig(d-mi)<0) mi=d;}}}}sort(p+mid-i,p+mid+j+1);return mi;}int main(){int i,n;while(scanf("%d",&n)!=EOF && n){for(i=0;i<n;i++) scanf("%lf%lf",&p[i].x,&p[i].y);sort(p,p+n);printf("%.2lf\n",min_dis(0,n)/2);}return 0;}