poj 3714 Raid 计算几何

题意:求两个集合中最近点对.

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#include<iostream> #include<cmath> #include<algorithm> #define eps 1e-10 using namespace std; //最近点对问题 const int maxn=200001; const double INF=1e100; typedef struct point { double x,y; int flag; point(){}; point(double xx,double yy){x=xx;y=yy;} }point; point p[maxn]; int n; int com(double x,double y) { if(x==y) return 0; if(x>y) return 1; return -1; } double min(double a, double b) { return a<b?a:b; } bool cmp1(point a,point b)//对于x进行排序 { return com(a.x,b.x)<0; } bool cmp2(int a,int b)//对于y进行排序 { return com(p[a].y,p[b].y)<0; } double dist(point &a,point &b)//传入两个点，求两点距离 { return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y)); } int b[maxn], k; double min_dis(point p[],int left,int right)//求从left到right这些点p[i]的最近点对 { int mid=(left+right)>>1,i,j; double ret=INF; if(left>=right) return ret; for(i=mid; i>=left && !com(p[i].x,p[mid].x); i--);//取中间的点 ret=min_dis(p,left,i); for(i=mid; i<=right && !com(p[i].x,p[mid].x); i++); ret=min(ret,min_dis(p,i,right)); k = 0;//寻找那些到我们所参考的点比ret小的点的个数 for(i=mid; i>=left && com(p[mid].x-p[i].x,ret)<=0; i--) b[++k]=i; for(i=mid+1; i<=right && com(p[i].x-p[mid].x,ret)<=0; i++) b[++k]=i; sort(b+1,b+k+1,cmp2); for(i=1; i<=k; i++) for(j=1; j+i<=k; j++) if(i+j<=k && com(p[b[i+j]].y-p[b[i]].y,ret) < 0 && p[b[i]].flag != p[b[i+j]].flag) ret=min(ret,dist(p[b[i]],p[b[i+j]])); return ret; } int main() { // freopen("in.txt","r",stdin); int i,j,k,test; scanf("%d",&test); while(test--) { scanf("%d",&n); for(i=1;i<=n;i++) { scanf("%lf%lf",&p[i].x,&p[i].y); p[i].flag=1; } for(i=1; i<=n; i++) { scanf("%lf %lf", &p[i+n].x, &p[i+n].y); p[i+n].flag = 2; } n<<=1; sort(p+1,p+n+1,cmp1); printf("%.3lf/n",min_dis(p,1,n)); } return 0; }