直接一起来看代码吧：
#include <iostream>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
using namespace std;
const int N = 100005;
const double MAX = 10e10;
const double eps = 0.00001;
struct Point
{
double x, y;
int index;
} ;
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);
int main()
{
int n, i;
int d;
cin>>n;
for (i = 0; i < n; i++)
{
cin>>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);
cout <<d<<endl;
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 m = (p + q) / 2;
int i, j, k;
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);
merge(b, c, p, m, q); //数组c左右部分分别是对y坐标有序的, 将其合并到b.
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) //青年人网提示求两个点的距离函数
{
double x1 = p.x - q.x, y1 = p.y - q.y;
return 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) //判断两点x坐标的大小关系
{
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) ////判断两点y坐标的大小关系
{
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) //返回两个double数的最小值
{
return (p > q) ? (q): (p);
}
责任编辑：小草
