HDU 1007 Quoit Design

给一系列点的坐标，求距离最小的两个点。

基本是照着别人的代码写的……我以为我看懂了的……怎么还是错了辣么多次……

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cmath>
 
using namespace std;
 
const int MAX = 100005;
 
struct point
{
    double x, y;
}p[MAX];
 
int pt[MAX];
 
bool cmpx(point a, point b)         //点按横坐标排序
{
    return a.x < b.x;           
}
 
bool cmpy(int a, int b)             //按纵坐标排序
{
    return p[a].y < p[b].y;         //卧槽我以前是这么写的return p[pt[a]].y < p[pt[b]].y;竟然曾经能ac
}
 
double dis(point a, point b)        //求两点距离
{
    return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));
}
 
double min_dis(int l, int r)        //求a[l]和a[r]之间的最短距离
{
    if (r - l == 1)
    {
        return dis(p[l], p[r]);
    }
    if (r - l == 2)                 //三个点时求三段距离的最小值
    {
        return min(min(dis(p[l], p[l + 1]), dis(p[l + 1], p[r])), dis(p[l], p[r]));
    }
    int mid = (l + r) >> 1;
    double ans = min(min_dis(l, mid), min_dis(mid + 1, r));    //把点按横坐标分为两部分，求左边的最小值和右边的最小值
    int cnt = 0;
    for (int i = l; i <= r; i++)    //两部分之间可能有最小值，只有横坐标之差小于ans时才有可能
    {
        if (p[i].x - p[mid].x <= ans && p[mid].x - p[i].x <= ans)
        {
            pt[cnt++] = i;
        }
    }
    sort(pt, pt + cnt, cmpy);
    for (int i = 0; i < cnt; i++)
    {
        for (int j = i + 1; j < cnt; j++)
        {
            if (p[pt[j]].y - p[pt[i]].y > ans)          //已经按纵坐标排序，这里加判断能节省时间
                break;
            ans = min(ans, dis(p[pt[j]], p[pt[i]]));
        }
    }
    return ans;
}
 
int main()
{
    int n;
    while (scanf("%d", &n) == 1 && n)
    {
        for (int i = 0; i < n; i++)
        {
            scanf("%lf%lf", &p[i].x, &p[i].y);
        }
        sort(p, p + n, cmpx);
        printf("%.2f\n", min_dis(0, n - 1) / 2);
    }
    return 0;
}