hdu1007 Quoit Design

平面最近点对板子,一个有趣的分治
还挺好写的...

#include <cstdio>
#include <cmath>
#include <vector>
#include <algorithm>
using namespace std;
const int maxn=100005;
typedef pair<double,double> pd;
pd A[maxn];
pd B[maxn];
pd C1[maxn],C2[maxn];
double sqr(double x){return x*x;}
double dis(pd a,pd b){
    return sqrt(sqr(a.first-b.first)+sqr(a.second-b.second));
}
double solve(int l,int r){
    if(l==r)return 1e20;
    int mid=(l+r)/2;
    double midx=(A[mid].first+A[mid+1].first)/2;
    double a1=solve(l,mid);
    double a2=solve(mid+1,r);
    double ans=min(a1,a2);
    int k1=0,k2=0;
    for(int i=l;i<=mid;++i){
        if(A[i].first+ans>=midx)C1[++k1]=A[i];
    }
    for(int i=mid+1;i<=r;++i){
        if(A[i].first-ans<=midx)C2[++k2]=A[i];
    }
    for(int i=1,L=1,R=0;i<=k1;++i){
        while(R<k2&&C1[i].second+ans>=C2[R+1].second)++R;
        while(L<=k2&&C1[i].second-ans>C2[L].second)++L;
        for(int j=L;j<=R;++j){
            ans = min(ans,dis(C1[i],C2[j]));
        }
    }
    for(int i=l,j=mid+1;i<=mid||j<=r;(j>r||(i<=mid&&A[i].second<=A[j].second))?(i++):(j++)){
        B[i+j-mid-1]=(j>r||(i<=mid&&A[i].second<=A[j].second))?(A[i]):(A[j]);
    }
    for(int i=l;i<=r;++i)A[i]=B[i];
    return ans;
}
int main(){
    int n;
    while(scanf("%d",&n),n!=0){
        for(int i=1;i<=n;++i){
            scanf("%lf%lf",&A[i].first,&A[i].second);
        }
        sort(A+1,A+n+1);
        printf("%.2fn",solve(1,n)/2.0);
    }
    return 0;
}