ACM模版

描述

题解

平面最近点对问题，模版题……反正本渣渣是直接套模版的，复杂度O(nlogn)。

等白天了好好看看算导，这个算法怎么搞的~~~

代码

#include <iostream>
#include <cmath>
#include <cstdio>

/* * O(N * logN) */
const int MAXN = 100005;
const double EPS = 0.00001;

struct Point
{
    double x, y;
    int index;
};

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

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);

int main()
{
    int n, i;
    double d;
    while (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)) / 2;

        printf("%.2lf\n", d);
    }

    return 0;
}

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);
}

参考

《平面最近点对》