HDU1007.Quoit Design

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题目大意 ：给你一堆点，求一个最小圆能够覆盖两个点的半径（最近两点距离的一半）；

最多100000个点，暴力即O(n^2)会超时，考虑二分，先求左边最短距离dl,右边dr, 和一个点左， 一个点在右的情况， 求d;

#include<bits/stdc++.h>

using namespace std;

#define lint long long
#define maxn 100005
#define mod 1e9+7

struct point
{
    double x, y;
};

int n;
point a[maxn];
vector<point> rp, lp;

bool cmpx(point a, point b)
{
    return a.x == b.x? a.y < b.y: a.x < b.x;
}

bool cmpy(point a, point b)
{
    return a.y == b.y? a.x < b.x: a.y < b.y;
}

double dis(point a ,point b)
{
    return sqrt(pow(a.x - b.x , )+pow(a.y - b.y, ));
}

double solver(int l, int r)
{
    if(r - l == ) return dis(a[l], a[r]);
    else if(l == r) return ;
    int mid = (l+r) >> ;
    double d = min(solver(l, mid), solver(mid+, r));
    lp.clear(); rp.clear();

    for(int i = l; i <= mid; i++)
        if(a[mid].x - a[i].x <= d)
        lp.push_back(a[i]);

    for(int i = mid+; i <= r; i++)
        if(a[i].x - a[mid].x < d)
        rp.push_back(a[i]);

    sort(rp.begin(), rp.end(), cmpy);
    for(int i = ; i < lp.size(); i++)
        for(int j = ; j < rp.size(); j++)
    {
        if(fabs(rp[j].y - lp[i].y) < d) d = min(dis(rp[j], lp[i]), d);
        if(rp[j].y - lp[i].y >= d) break;
    }

    return d;

}

int main()
{
    int n;
    cin >> n;
    while(n)
    {
        for(int i = ; i <= n; i++)
            scanf("%lf%lf", &a[i].x,&a[i].y);
        sort(a+, a+n+, cmpx);
        double d = solver(, n)/;
        printf("%.2lf\n", d);
        cin >> n;
    }

}