poj 3714 最近点对

poj 3714 最近点对
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

/*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;}