求平面上最近点对 杭电1007

 

#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;

static const int MAXNUM = 100000;

struct Point{
    double x,y;
}arrP[MAXNUM],arrX[MAXNUM],arrY[MAXNUM];

bool CompX(Point a, Point b)
{
    if(a.x == b.x)
        return a.y < b.y;
    return a.x < b.x;
}

bool CompY(Point a, Point b)
{
    if(a.y == b.y)
        return a.x < b.x;
    return a.y < b.y;
}

double GetDistance(Point a, Point b)
{
    return sqrt(pow(a.x - b.x, 2.0) + pow(a.y - b.y,2.0));
}

double MinDistance(double a, double b)
{
    return a<=b? a: b;
}

double GetMinDistabce(int l, int r)
{
    if(l == r)
        return 0;
    else if(l+1 == r)
        return GetDistance(arrP[l], arrP[r]);
    else if(l+2 == r){
        double tmp1 = GetDistance(arrP[l],arrP[l+1]);
        double tmp2 = GetDistance(arrP[l],arrP[r]);
        double tmp3 = GetDistance(arrP[l+1],arrP[r]);
        return MinDistance(MinDistance(tmp1,tmp2),tmp3);
    }


    int middle=(l+r)/2,j=0,k=0;

    double minDistance = MinDistance(GetMinDistabce(l,middle),GetMinDistabce(middle+1,r));
    for(int i=l;i<=r;i++){
        if(fabs(arrP[i].x - arrP[middle].x) < minDistance)
            arrX[j++] = arrP[i];//arrX[]存储的是x坐标距离小于当前最小距离的点
    }
    sort(arrX,arrX+j,CompY);
    middle = (j+1)/2;
    for(int i = 0;i < j; i++){
        if(fabs(arrX[i].y - arrX[middle].y) < minDistance)
            arrY[k++]=arrX[i];//arrY[]存储的是x坐标距离小于当前最小距离的点，且Y坐标距离也小于当前最小点的点
    }
    for(int i=0; i<k; i++){//遍历arrY数组，两两计算点之间距离
        for(int j=i+1; j<k; j++){
            double tmpDistance = GetDistance(arrY[i],arrY[j]);
            if(tmpDistance < minDistance)
                minDistance = tmpDistance;
        }
    }
    return minDistance;
}

int main()
{
    int n;
    while(cin>>n && n!=0){
        //Point p;
        double minDistance;
        for(int i=0; i<n; i++){
            cin >> arrP[i].x >> arrP[i].y;
        }
        sort(arrP,arrP+n,CompX);
        minDistance = GetMinDistabce(0,n-1);
        cout.precision(2);
        cout << fixed << minDistance/2<<endl;
    }
    return 0;
}