/*poj 3714 题意: 给出n个a类点,n个b类点,求a类点到b类点的最近距离。 限制: 1 <= n <= 1e5 0 <= x,y <= 1e9 思路: 点分治 */ #include<iostream> #include<cstdio> #include<algorithm> #include<cmath> using namespace std; const double INF=1e30; struct Point{ double x,y; int s; Point(){} Point(double _x,double _y){ x=_x; y=_y; } Point operator - (Point p){ return Point(x-p.x,y-p.y); } double dot(Point p){ return x*p.x+y*p.y; } }; const int N=1e5+5; Point p[2*N]; int t[2*N]; bool cmpxy(Point a,Point b){ if(a.x==b.x) return a.y<b.y; return a.x<b.x; } bool cmpy(int a,int b){ return p[a].y<p[b].y; } double dist(Point a,Point b){ return sqrt((a-b).dot(a-b)); } double gao(int l,int r){ double ret=INF; if(l==r) return ret; if(l+1==r){ if(p[l].s!=p[r].s) return dist(p[l],p[r]); else return ret; } int mid=(l+r)>>1; double d=min(gao(l,mid),gao(mid+1,r)); int cnt=0; for(int i=l;i<r;++i){ if(fabs(p[mid].x-p[i].x)<d) t[cnt++]=i; } sort(t,t+cnt,cmpy); for(int i=0;i<cnt;++i) for(int j=i+1;j<cnt && p[t[j]].y-p[t[i]].y<d;++j) if(p[t[i]].s!=p[t[j]].s) d=min(d,dist(p[t[i]],p[t[j]])); return d; } int main(){ int T; scanf("%d",&T); int n; while(T--){ scanf("%d",&n); for(int i=0;i<n;++i){ scanf("%lf%lf",&p[i].x,&p[i].y); p[i].s=0; } for(int i=n;i<2*n;++i){ scanf("%lf%lf",&p[i].x,&p[i].y); p[i].s=1; } sort(p,p+2*n,cmpxy); printf("%.3f\n",gao(0,2*n-1)); } return 0; }
原文地址:http://blog.csdn.net/whai362/article/details/44275665