UVA 10245 The Closest Pair Problem

n=1的情况要注意~

#include<cstdio>#include<iostream>#include<algorithm>#include<cstring>#include<cmath>using namespace std;const int maxn = 10060;// 分治算法求最近点对struct point//保存每一个点{    double x,y;}p[maxn];int a[maxn];//保存筛选的坐标点的索引即2min(dl,dh)区间的坐标点索引int cmpx(point a,point b)//对n个点按横坐标由小到大排序{    return a.x<b.x;}int cmpy(int a,int b)//（这里用的是下标索引）对筛选的点按纵坐标由小到大排序{    return p[a].y<p[b].y;}inline double dis(point a,point b)//求点对间的距离{    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));}double closest(int low,int high)//求最近点对{    int i,j,k;    if(low+1==high)//只有两个点        return dis(p[low],p[high]);    if(low+2==high)//只有三个点        return min(dis(p[low],p[high]),min(dis(p[low],p[low+1]),dis(p[low+1],p[high])));    int mid=(low+high)>>1;//求中点即左右子集的分界线    double d=min(closest(low,mid),closest(mid+1,high));    for(i=low,k=0;i<=high;i++)//把x坐标在p[mid].x-d  ~  p[mid].x+d范围内的点筛选出来    {        if(p[i].x >=p[mid].x-d&&p[i].x<=p[mid].x+d)            a[k++]=i;//保存这些点的下标索引    }    sort(a,a+k,cmpy);//按y坐标进行升序排序    for(i=0;i<k;i++)    {        for(j=i+1;j<k;j++)        {            if(p[a[j]].y-p[a[i]].y>=d)//注意下标索引                break;            d=min(d,dis(p[a[i]],p[a[j]]));        }    }    return d;}int main(){    int i,n;    while(scanf("%d",&n) &&n!=0)//n个点    {        for(i = 0 ; i < n ; ++i)            scanf("%lf %lf",&p[i].x,&p[i].y);        if(n==1) printf("INFINITY\n");        else        {            sort(p , p + n , cmpx);//按x坐标进行升序排序            if(closest(0 , n - 1)>10000) printf("INFINITY\n");            else printf("%.4f\n",closest(0 , n - 1));//最近点对间的距离        }    }    return 0;}