題意:
給你平面上的一些點,讓你找出這些點的最近點對的距離
題解:
採用分治,達到O(nlognlogn)的時間複雜度就能艹過去了
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<math.h>
using namespace std;
//O(nlognlogn)找最近點對
const int N=1e5+7;
struct TPoint
{
double x,y;
}ply[N],ans[N];
int n;
inline double MIN(double a,double b) {return a<b?a:b;}
bool cmpx(TPoint a,TPoint b) {return a.x<b.x;}
bool cmpy(TPoint a,TPoint b) {return a.y<b.y;}
double dist(TPoint a,TPoint b)
{
double s1=a.x-b.x;
double t1=a.y-b.y;
return sqrt(s1*s1+t1*t1);
}
double closest(int l,int r)
{
if(l+1==r) return dist(ply[l],ply[r]);//2點
else if(l+2==r)//三點
return MIN(dist(ply[l],ply[l+1]),MIN(dist(ply[l],ply[r]),dist(ply[l+1],ply[r])));
int i,j,mid,cnt;
mid=(l+r)>>1;
double mi=MIN(closest(l,mid),closest(mid+1,r));//遞歸解決
for(i=l,cnt=0;i<=r;i++)//相鄰點符合
{
if(fabs(ply[i].x-ply[mid].x)<=mi)
ans[cnt++]=ply[i];
}
sort(ans,ans+cnt,cmpy);//按y排序
for(i=0;i<cnt;i++)for(j=i+1;j<cnt;j++)//更新最小距離
{
if(ans[j].y-ans[i].y>=mi) break;
mi=MIN(mi,dist(ans[i],ans[j]));
}
return mi;
}
int main()
{
while(scanf("%d",&n),n)
{
int i;
for(i=0;i<n;i++) scanf("%lf%lf",&ply[i].x,&ply[i].y);//輸入點
sort(ply,ply+n,cmpx);//按x排序
double mi=closest(0,n-1);
printf("%.2lf\n",mi/2);
}
return 0;
}