HUD 1007 最近点对(分治法)

题解：

给了近100000个点，要求最近点对，暴力显然不现实，那么就用分治法去做。

最近点对，参照：

http://www.cnblogs.com/AdaByron/archive/2011/10/07/2200966.html

套用吉大模板：

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cstring>
using namespace std;

const int N = 100005;
const double MAX = 10e100, eps = 0.00001;

typedef struct TagPoint {
    double x;
    double y;
    int index;
} Point;

Point a[N], b[N], c[N];

double closest(Point *, Point *, Point *, int, int);
double dis(Point, Point);
int cmp_x(const void *, const void*);
int cmp_y(const void *, const void*);
int merge(Point *, Point *, int, int, int);
inline double min(double, double);

double closest(Point a[],Point b[],Point c[],int p,int q) {
    if (q - p == 1) return dis(a[p], a[q]);
    if (q - p == 2) {
        double x1 = dis(a[p], a[q]);
        double x2 = dis(a[p + 1], a[q]);
        double x3 = dis(a[p], a[p + 1]);
        if (x1 < x2 && x1 < x3) return x1;
        else if (x2 < x3) return x2;
        else return x3;
    }
    int i, j, k, m = (p + q) / 2;
    double d1, d2;
    for (i = p, j = p, k = m + 1; i <= q; i++)
        if (b[i].index <= m) c[j++] = b[i];
    //数组c左半部保存划分后左部的点, 且对y是有序的.
        else c[k++] = b[i];
    d1 = closest(a, c, b, p, m);
    d2 = closest(a, c, b, m + 1, q);
    double dm = min(d1, d2);
    //数组c左右部分分别是对y坐标有序的, 将其合并到b.
    merge(b, c, p, m, q);
    for (i = p, k = p; i <= q; i++)
        if (fabs(b[i].x - b[m].x) < dm) c[k++] = b[i];
    //找出离划分基准左右不超过dm的部分, 且仍然对y坐标有序.
    for (i = p; i < k; i++)
        for (j = i + 1; j < k && c[j].y - c[i].y < dm; j++) {
            double temp = dis(c[i], c[j]);
            if (temp < dm) dm = temp;
        }
    return dm;
}
double dis(Point p, Point q) {
    double x1 = p.x - q.x, y1 = p.y - q.y;
    return sqrt(x1 *x1 + y1 * y1);
}

int merge(Point p[], Point q[], int s, int m, int t) {
    int i, j, k;
    for (i=s, j=m+1, k = s; i <= m && j <= t;) {
        if (q[i].y > q[j].y)
            p[k++] = q[j], j++;
        else
            p[k++] = q[i], i++;
    }
    while (i <= m) p[k++] = q[i++];
    while (j <= t) p[k++] = q[j++];
    memcpy(q + s, p + s, (t - s + 1) *sizeof(p[0]));
    return 0;
}

int cmp_x(const void *p, const void *q) {
    double temp = ((Point*)p)->x - ((Point*)q)->x;
    if (temp > 0)
        return 1;
    else if (fabs(temp) < eps)
        return 0;
    else
        return - 1;
}

int cmp_y(const void *p, const void *q) {
    double temp = ((Point*)p)->y - ((Point*)q)->y;
    if(temp > 0)
        return 1;
    else if (fabs(temp) < eps)
        return 0;
    else
        return - 1;
}
inline double min(double p, double q) {
    return (p > q) ? (q): (p);
}

int main() {
    int n, i;
    double d;
    while(EOF != scanf("%d", &n), n) {
        for (i = 0; i < n; ++i) {
            scanf("%lf %lf", &a[i].x, &a[i].y);
        }
        qsort(a, n, sizeof(a[0]), cmp_x);
        for (i = 0; i < n; i++) {
            a[i].index = i;
        }
        memcpy(b, a, n * sizeof(a[0]));
        qsort(b, n, sizeof(b[0]), cmp_y);
        d = closest(a, b, c, 0, n - 1);
        printf("%.2f\n", d / 2.0);
    }
    return 0;
}