题意：
给你平面上的一些点，让你找出这些点的最近点对的距离
题解：
采用分治，达到O(nlognlogn)的时间复杂度就能艹过去了
1 #include<stdio.h> 2 #include<string.h> 3 #include<algorithm> 4 #include<math.h> 5 using namespace std; 6 //O(nlognlogn)找最近点对 7 const int N=1e5+7; 8 struct TPoint 9 { 10 double x,y; 11 }ply[N],ans[N]; 12 int n; 13 inline double MIN(double a,double b) {return a<b?a:b;} 14 bool cmpx(TPoint a,TPoint b) {return a.x<b.x;} 15 bool cmpy(TPoint a,TPoint b) {return a.y<b.y;} 16 double dist(TPoint a,TPoint b) 17 { 18 double s1=a.x-b.x; 19 double t1=a.y-b.y; 20 return sqrt(s1*s1+t1*t1); 21 } 22 double closest(int l,int r) 23 { 24 if(l+1==r) return dist(ply[l],ply[r]);//2点 25 else if(l+2==r)//三点 26 return MIN(dist(ply[l],ply[l+1]),MIN(dist(ply[l],ply[r]),dist(ply[l+1],ply[r]))); 27 int i,j,mid,cnt; 28 mid=(l+r)>>1; 29 double mi=MIN(closest(l,mid),closest(mid+1,r));//递归解决 30 for(i=l,cnt=0;i<=r;i++)//相邻点符合 31 { 32 if(fabs(ply[i].x-ply[mid].x)<=mi) 33 ans[cnt++]=ply[i]; 34 } 35 sort(ans,ans+cnt,cmpy);//按y排序 36 for(i=0;i<cnt;i++)for(j=i+1;j<cnt;j++)//更新最小距离 37 { 38 if(ans[j].y-ans[i].y>=mi) break; 39 mi=MIN(mi,dist(ans[i],ans[j])); 40 } 41 return mi; 42 } 43 int main() 44 { 45 while(scanf("%d",&n),n) 46 { 47 int i; 48 for(i=0;i<n;i++) scanf("%lf%lf",&ply[i].x,&ply[i].y);//输入点 49 sort(ply,ply+n,cmpx);//按x排序 50 double mi=closest(0,n-1); 51 printf("%.2lf ",mi/2); 52 } 53 return 0; 54 }