hdu 1007 Quoit Design (Nearest Point Pair)

http://acm.hdu.edu.cn/showproblem.php?pid=1007

　　最近点对。

　　做法是先对所有点先按x再按y排序，然后进行分治搜索最近点对。对于每个区间，如果点数小于3，就直接暴力搜索最近点，否则对其进行分治。分治出来的两个区间，我们要挑选出与中点距离小于已找到的最近距离的所有点，然后对他们进行暴力枚举最近距离。根据《算法导论》证明的，合并两个区间之后，被挑选出来的点不会超过6个，于是可以得到总的时间复杂度是O(nlogn)。

代码如下：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <cmath>

using namespace std;

const int N = 111111;
const double FINF = 1e100;

template <class T> T sqr(T x) { return x * x;}
struct Point {
    double x, y;
    Point() {}
    Point(int x, int y) : x(x), y(y) {}
    bool operator < (Point a) const {
        return x < a.x || x == a.x && y < a.y;
    }
    Point operator + (Point a) {
        return Point(x + a.x, y + a.y);
    }
} p[N];

bool cmpY(int x, int y) { return p[x].y < p[y].y;}
inline double dist(Point x, Point y) { return sqrt(sqr(x.x - y.x) + sqr(x.y - y.y));}

#define lson l, m
#define rson m + 1, r

int arr[N];

double closest(int l, int r) {
    if (l >= r) return FINF;
    if (l + 1 == r) return dist(p[l], p[r]);
    int m = l + r >> 1;
    double cd = min(closest(lson), closest(rson));
    int end = 0;
    for (int i = l; i <= r; i++) {
        if (dist(p[i], p[m]) <= cd) {
            arr[end++] = i;
        }
    }
    sort(arr, arr + end, cmpY);
    for (int i = 0; i < end; i++) {
        for (int j = i + 1; j < end; j++) {
            cd = min(cd, dist(p[arr[i]], p[arr[j]]));
        }
    }
    return cd;
}

int main() {
//    freopen("in", "r", stdin);
    int n;
    while (~scanf("%d", &n) && n) {
        for (int i = 0; i < n; i++) {
            scanf("%lf%lf", &p[i].x, &p[i].y);
        }
        sort(p, p + n);
        printf("%.2f\n", closest(0, n - 1) / 2.0);
    }
    return 0;
}

UPD：

　　之前的代码估计是水过了，更新一下代码：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <cmath>

using namespace std;

template<class T> T sqr(T x) { return x * x;}
typedef pair<double, double> Point;
#define x first
#define y second
inline double dist(Point a, Point b) { return sqrt(sqr(a.x - b.x) + sqr(a.y - b.y));}
inline bool cmpx(Point a, Point b) { return a.x < b.x;}
inline bool cmpy(Point a, Point b) { return a.y < b.y;}
const int N = 111111;
const double FINF = 1e100;

#define lson l, m
#define rson m + 1, r
Point pt[N], tmp[N];

double npp(int l, int r) {
    if (r - l <= 3) {
        double ret = FINF;
        for (int i = l; i < r; i++) {
            for (int j = i + 1; j <= r; j++) {
                ret = min(ret, dist(pt[i], pt[j]));
            }
        }
        return ret;
    }
    int m = l + r >> 1;
    Point mid = pt[m];
    double md = min(npp(lson), npp(rson));
    int tt = 0;
    for (int i = l; i <= r; i++) if (fabs(pt[i].x - mid.x) <= md) tmp[tt++] = pt[i];
    sort(tmp, tmp + tt, cmpy);
    for (int i = 0; i < tt; i++) {
        int j = i + 1;
        while (j < tt) {
            if (tmp[j].y - tmp[i].y > md) break;
            md = min(md, dist(tmp[i], tmp[j]));
            j++;
        }
    }
    return md;
}

int main() {
    //freopen("in", "r", stdin);
    int n;
    while (~scanf("%d", &n) && n) {
        for (int i = 0; i < n; i++) scanf("%lf%lf", &pt[i].x, &pt[i].y);
        sort(pt, pt + n, cmpx);
        printf("%.2f\n", npp(0, n - 1) / 2);
    }
    return 0;
}

——written by Lyon